Solving Trigonometric Equations - Trigonometry

Card 0 of 264

Question

Solve for : $0 = 3 \cos ^2 x + 8 \cos x + 4$

Answer

Use the quadratic formula:

$\cos x = \frac{ -8 \pm \sqrt{64 - 4(3)(4)}}{2(3)} = \frac{-8 \pm \sqrt{16}}{6 } = \frac{-8\pm4}{6}$

$\cos x = -2 \enspace or - \frac{2}{3}$

-2 is outside the range of cosine, so the answer has to come from $- \frac{2}{3}$ :

$\cos x = - \frac{2}{3}$

$x = \cos^{-1} (-\frac{2}{3} ) \approx 131.81^o$ according to a calculator

The other angle with a cosine of $-\frac{2}{3}$ is

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Question

In the interval $0\leq x\leq\pi$ , what values of x satisfy the following equation?

$1-\sin x=\cos^2x-\frac{1}{4}$

Answer

We start by rewriting the $\cos^2x$ term on the right hand side in terms of $\sin^2x$ .

$\cos^2x=1-\sin^2x$

$1-\sin x=1-\sin^2x-\frac{1}{4}$

We then move everything to the left hand side of the equation and cancel.

$1-\sin x-1+\sin^2x+\frac{1}{4}=0$

$\sin^2x-\sin x+\frac{1}{4}=0$

Apply the quadratic formula:

$\begin{align} \sin x&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(\frac{1}{4})}}{2(1)}\ &=\frac{1\pm\sqrt{1-1}}{2}\ &=\frac{1}{2} \end{align}$

So $\sin x=\frac{1}{2}$ . Using the unit circle, the two values of that yield this are $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$ .

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Question

Solve the equation

for $0\leq x\leq2\pi$ .

Answer

First of all, we can use the Pythagorean identity to rewrite the given equation in terms of .

$\Rightarrow 2(1-sin^2x)+sinx=1$

$\Rightarrow 2-2sin^2x+sinx=1$

$\Rightarrow -2sin^2x+sinx+1=0$

$\Rightarrow 2sin^2x-sinx-1=0$

This is a quadratic equation in terms of ; hence, we can use the quadratic formula to solve this equation for .

$sinx=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where .

$\Rightarrow sinx=\frac{1\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

$\Rightarrow sinx=\frac{1\pm3}{4}$

$\Rightarrow sinx=-\frac{1}{2}, sinx=1$ .

Now, when $x=\frac{\pi}{2}$ , and $sinx=-\frac{1}{2}$ when $x=\frac{7\pi}{6}$ or $\frac{11\pi}{6}$ .

Hence, the solutions to the original equation are

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

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Question

Solve the following system:

Answer

A number x is a solution if it satisfies both equations.

We note first we can write the first equation in the form :

$cosx\ge \frac{4}{3}$

We know that $cosx\le 1$ for all reals. This means that there is no x that

satisifies the first inequality. This shows that the system cannot satisfy both equations since it does not satisfy one of them. This shows that our system does not have a solution.

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Question

Solve for $\theta$ : $2 \sin ^2 (3 \theta ) = 1$

Answer

There are multiple solution paths. We could subtract 1 from both sides and use the quadratic formula with and . Or we could solve using inverse opperations:

$2 \sin ^2 (3 \theta) = 1$ divide both sides by 2

$\sin ^2 (3 \theta ) = \frac{1}{2}$ take the square root of both sides

$\sin (3 \theta ) = \pm \frac{1} { \sqrt 2 }$

The unit circle tells us that potential solutions for $3 \theta$ are $\frac{\pi }{4}, \frac{3 \pi }{4} , \frac{5 \pi }{4} , \frac{7 \pi }{4}$ .

To get our final solution set, divide each by 3, giving:

$\frac{ \pi }{12}, \frac{\pi }{4 }, \frac{5 \pi }{12}, \frac{ 7 \pi }{12 }$ .

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Question

Solve for $\theta$ : $4 \sin ^2( \theta + \frac{ \pi }{3 }) + 2 = 5$

Answer

This problem has multiple solution paths, including subtracting 5 from both sides and using the quadratic formula with . We can also solve using inverse opperations:

$4 \sin ^2 (\theta + \frac{ \pi }{3} ) + 2 = 5$ subtract 2 from both sides

$4 \sin ^2 (\theta + \frac{ \pi }{3} ) = 3$ divide both sides by 4

$\sin ^ 2 (\theta + \frac{ \pi }{3} ) = \frac{3}{4}$ take the square root of both sides

$\sin (\theta + \frac{ \pi }{3} ) = \pm \frac{\sqrt3}{2}$

If the sine of an angle is $\pm \frac{\sqrt 3 }{2}$ , that angle must be one of $\frac{ \pi }{3} , \frac{2 \pi }{3} , \frac{4 \pi }{3} , \frac{5 \pi }{3}$ . Since the angle is $\theta + \frac{ \pi }{3}$ , we can get theta by subtracting $\frac{ \pi }{3}$ :

$\frac{\pi }{3} - \frac{ \pi }{3} = 0$

$\frac { 2 \pi }{3} - \frac{ \pi }{3} = \frac{ \pi }{3}$

$\frac{ 4 \pi }{3} - \frac{ \pi }{3} = \frac{ 3 \pi }{3} = \pi$

$\frac{ 5 \pi }{3} - \frac{ \pi }{3} = \frac{ 4 \pi }{3}$

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Question

For this question, we will denote by max the maximum value of the function and min the minimum value of the function.

What is the maximum and minimum values of

where is a real number.

Answer

To find the maximum and the minimum , we can view the above function as

a system where $-1\le cosx\le 1$ and $-1\le sinx\le 1$ . Using these two conditions we find the maximum and the minimum.

$0\le cos^2x\le 1$ means also that $0\le cos^4x\le 1$ ( $\star$ ) We also have:

$-1\le sinx\le 1$ implies that :

$-1\le sin^{3}x\le 1$ ( $\star \star$ ) Therefore we have by adding ( $\star$ ) and( $\star \star$ )

$-1\le cos^4x+sin^3x\le 2$

This means that max=2 and min=-1

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Question

Solve the following equation for $0\leq x<2\pi$ .

$\sin\left(\frac{x}{2}\right)=\cos x -1$

Answer

$\sin\left(\frac{x}{2}\right)=\cos x -1$ ; We start by substituting a new variable. Let $u=\frac{x}{2}$ .

$\sin u=\cos 2u-1$ ; Use the double angle identity for cosine

$\sin u = 1-2\sin ^2 u-1$ ; the 1's cancel, so add $2\sin ^2u$ to both sides

$2\sin ^2u+\sin u=0$ ; factor out a $\sin u$ from both terms.

$\sin u(2\sin u+1)=0$ ; set each expression equal to 0.

$\sin u =0$ or $2\sin u +1=0$ ; solve the second equation for sin u.

$\sin u = 0$ or $\sin u = -\frac{1}{2}$ ; take the inverse sine to solve for u (use a unit circle diagram or a calculator)

$\frac{x}{2}=u=0, \frac{7\pi}{6}, \frac{11\pi}{6}$ ; multiply everything by 2 to solve for x.

$x=0, \frac{7\pi}{3}, \frac{11\pi}{3}$ ; Notice that the last two solutions are not within our range $0\leq x<2\pi$ . So the only solution is .

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Question

Solve the equation below for greater than or equal to and strictly less than .

Answer

Recall the values of for which . If it helps, think of sine as the values on the unit circle. Thus, the acceptable values of would be 0, 180, 360, 540 etc.. However, in our scenario .

Thus we have and .

Any other answer would give us values greater than 90. When we divide by 4, we get our answers,

and .

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Question

$f(x)=2\sin{(2x)}-2\sin(x)\cos(x)$

Find the three smallest positive roots of the above equation.

Answer

By the double angle identity, we can find

$-2\sin(x)\cos(x)=-\sin(2x)$

$f(x)=2\sin(2x)-\sin(2x)=\sin(2x)$

So to get the zeros, solve:

$\sin(2x)=0$

This means that any number that when doubled equals a multiple of 180 degrees is a zero. In this case that includes

$0, \pm 90^{\circ}, \pm180^{\circ}, \pm270^{\circ}, etc.$

But the question asks for the smallest positive roots which excludes the negative and zero roots, leaving 90, 180, 270

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Question

Which of the following is NOT a solution to the equation below such that $0\leq x<2\pi$ ?

Answer

Given the multiple choice nature of the problem, the easiest way to solve would be to simply plug in each answer and find the one that does not work.

However, we want to learn the math within the problem. We begin solving the equation by factoring

We then divide.

$cos^2(x)-sin^2(x)=\frac{1}{2}$

We then must remember that our left side is equivalent to something simpler.

We can therefore substitute.

$cos(2x)=\frac{1}{2}$

We then must consider the angles whose cosine is $\frac{1}{2}$ . The two angles within the first revolution of the unit circle are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ , but since our angle is , we need to consider the second revolution, which also gives us $\frac{7\pi}{3}$ and $\frac{11\pi}{3}$ .

But since is equal to each of these angles, we must divide them by 2 to find our answers. Therefore, we have

$x=\frac{\pi}{6},\frac{\5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}$

Therefore, there is only one answer choice that does not belong.

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Question

Which of the following is a solution to the following equation such that $0\leq x<2\pi?$

Answer

We begin by getting the right side of the equation to equal zero.

Next we factor.

We then set each factor equal to zero and solve.

or

$sin(x)=-\frac{1}{2}$

We then determine the angles that satisfy each solution within one revolution.

The angles $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ satisfy the first, and $\frac{\pi}{2}$ satisfies the second. Only $\frac{11\pi}{6}$ is among our answer choices.

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Question

Solve the following equation. Find all solutions such that $0\leq x <2\pi$ .

$2\sin x = 1$

Answer

$2\sin x=1$ ; Divide both sides by 2 to get

$\sin x=\frac{1}{2}$ ; take the inverse sine on both sides

$\sin^{-1}(\sin x)=\sin ^{-1}\bigg(\frac{1}{2}\bigg)$ ; the left side reduces to x, so

$x=\sin ^{-1}\bigg(\frac{1}{2}\bigg)$

At this point, either use a unit circle diagram or a calculator to find the value.

Keep in mind that the problem asks for all solutions between and $2\pi$ .

If you use a calculator, you will only get $\frac{\pi}{6}$ as an answer.

So we need to find another angle that satisfies the equation $\sin x=\frac{1}{2}$ .

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Question

Solve the following equation. Find all solutions such that $0\leq x<2\pi$ .

$\sin 2x=4\cos x$

Answer

$\sin 2x=4\cos x$ ; First use the double angle identity for $\sin 2x$ .

$2\sin x \cos x = 4\cos x$ ; divide both sides by 2

$\sin x \cos x=2\cos x$ ; subtract the $2\cos x$ from both sides

$\sin x \cos x-2\cos x=0$ ; factor out the $\cos x$

$\cos x(\sin x-2)=0$ ; Now we have the product of two expressions is 0. This can only happen if one (or both) expressions are equal to 0. So let each expression equal 0.

$\cos x= 0$ or $\sin x-2=0$ ;

$\cos x = 0$ or $\sin x=2$ ; Take the inverse of each function for each expression.

$x=\cos ^{-1}(0)$ or $x=\sin ^{-1}(2)$ ; The second equation is not possible so gives no solution, but the first equation gives us:

$x=\frac{\pi}{2}; \frac{3\pi}{2}$

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Question

Solve the following equation for $0\leq x<2\pi$ .

$\sin 2x=\cos 2x$

Answer

The fastest way to solve this problem is to substitute a new variable. Let .

The equation now becomes:

$\sin u=\cos u$

So at what angles are the sine and cosine functions equal. This occurs at

$u=\frac{\pi}{4}; \frac{5\pi}{4}; \frac{9\pi}{4}; \frac{13\pi}{4}$

You may be wondering, "Why did you include

$\frac{9\pi}{4}; \frac{13\pi}{4}$ if they're not between and $2\pi$ ?"

The reason is because once we substitute back the original variable, we will have to divide by 2. This dividing by 2 will bring the last two answers within our range.

$2x=\frac{\pi}{4}; 2x=\frac{5\pi}{4}; 2x=\frac{9\pi}{4}; 2x=\frac{13\pi}{4}$

Dividing each answer by 2 gives us

$x=\frac{\pi}{8}; \frac{5\pi}{8}; \frac{9\pi}{8}; \frac{13\pi}{8}$

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Question

Solve the equation for $0\leq x<2\pi$ .

$\cos 4x = \cos 2x$

Answer

We begin by substituting a new variable .

$\cos 2u=\cos u$ ; Use the double angle identity for $\cos 2u$ .

$2\cos ^2 u-1=\cos u$ ; subtract the $\cos u$ from both sides.

$2\cos ^2 u-\cos u-1=0$ ; This expression can be factored.

$(2\cos u+1)(\cos u-1)=0$ ; set each expression equal to 0.

$2\cos u+1=0$ or $\cos u -1=0$ ; solve each equation for $\cos u$

$\cos u = -\frac{1}{2}$ or $\cos u = 1$ ; Since we sustituted a new variable we can see that if $0\leq x<2\pi$ , then we must have $0\leq 2x < 4\pi$ . Since , that means $0\leq u<4\pi$ .

This is important information because it tells us that when we solve both equations for u, our answers can go all the way up to $4\pi$ not just $2\pi$ .

So we get

$2x = u=0; \frac{2\pi}{3}; \frac{4\pi}{3}; 2\pi; \frac{8\pi}{3}; \frac{10\pi}{3};$ Divide everything by 2 to get our final solutions

$x=0; \frac{\pi}{3}; \frac{2\pi}{3}; \pi; \frac{4\pi}{3}; \frac{5\pi}{3};$

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Question

Solve the following equation for $0\leq x<2\pi$ .

$4\cos ^2x=3$

Answer

$4\cos ^2x=3$ ; First divide both sides of the equation by 4

$\cos ^2x=\frac{3}{4}$ ; Next take the square root on both sides. Be careful. Remember that when YOU take a square root to solve an equation, the answer could be positive or negative. (If the square root was already a part of the equation, it usually only requires the positive square root. For example, the solutions to are 2 and -2, but if we plug in 4 into the function $f(x)=\sqrt{x}$ the answer is only 2.) So,

$\cos x = \pm \frac{\sqrt{3}}{2}$ ; we can separate this into two equations

$\cos x = \frac{\sqrt{3}}{2}$ and $\cos x =- \frac{\sqrt{3}}{2}$ ; we get

$x=\frac{\pi}{6}; \frac{11\pi}{6}$ and $x=\frac{5\pi}{6}; \frac{7\pi}{6}$

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Question

Solve the equation for $0\leq x<2\pi$ .

$3\tan^2x=1$

Answer

$3\tan^2x=1$ ; Divide both sides by 3

$\tan^2x=\frac{1}{3}$ ; Take the square root on both sides. Just as the previous question, when you take a square root the answer could be positive or negative.

$\tan x=\pm \frac{1}{\sqrt{3}}$ ; This can be written as two separate equations

$\tan x=\frac{1}{\sqrt{3}}$ and $\tan x=-\frac{1}{\sqrt{3}}$ ; Take the inverse tangent

$x=\frac{\pi}{6}; \frac{7\pi}{6}$ and $x=\frac{5\pi}{6}; \frac{11\pi}{6}$

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Question

Solve the following equation for $0\leq x<2\pi$ .

$\tan ^2x+2\tan x+1=0$

Answer

$\tan ^2x+2\tan x+1=0$ ; The expression is similar to a quadratic expression and can be factored.

$(\tan x+1)(\tan x+1)=0$ ; set both expressions equal to 0. Since they are the same, the solutions will repeat, so I will only write it once.

$\tan x+1=0$

$\tan x=-1$ ; take the inverse tangent on both sides

$x=\tan ^{-1}(-1)$

$x=\frac{3\pi}{4}; \frac{7\pi}{4}$

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Question

Solve the following equation for $0\leq x<2\pi$ .

$2\sin ^2x=3\cos(2x)$

Answer

$2\sin ^2x=3\cos(2x)$ ; use the double angle identity for cosine

$2\sin ^2x=3(1-2\sin ^2x)$ ; distribute the 3 on the right side

$2\sin ^2x=3-6\sin ^2x$ ; add the $6\sin ^2x$ to both sides

$8\sin ^2x=3$ ; divide both sides by 8

$\sin ^2x=\frac{3}{8}$ ; take the square root on both sides (Remember: it could be positive or negative)

$\sin x=\pm \frac{3}{8}$ ; separate into two equations and take the inverse sine

$x=\sin ^{-1}\left(\sqrt{\frac{3}{8}} \right )$ and $x=\sin ^{-1}\left(-\sqrt{\frac{3}{8}} \right )$ ; Use a calculator

and (calculator will give -0.659, but that is not in our range, so add $2\pi$ to get 5.624)

The last two solutions are found using a unit circle. Since our x can be negative or positive this means that there is a corresponding value in all of the quadrants.

Similarly, we can get our last answer x= 3.801 as this is the x value in the third quadrant and is found by adding $0.659+\pi=3.801$ .

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