Solving Trigonometric Equations

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Trigonometry › Solving Trigonometric Equations

Questions 1 - 10

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

$\frac{11\pi}{6}$

$\frac{\pi}{6}$

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

$\frac{5\pi}{4}$

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

Explanation

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.

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

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.

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

$\frac{11\pi}{6}$

$\frac{\pi}{6}$

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

$\frac{5\pi}{4}$

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

Explanation

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.

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

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.

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

$\sin 2x=\cos 2x$

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

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

$x=\frac{\pi}{4}; \frac{5\pi}{4}$

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

No solution exists

Explanation

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

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

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

$\sin 2x=\cos 2x$

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

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

$x=\frac{\pi}{4}; \frac{5\pi}{4}$

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

No solution exists

Explanation

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

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

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

$\cos 4x = \cos 2x$

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

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

$x=0; \frac{2\pi}{3}; \frac{4\pi}{3}; 2\pi; \frac{8\pi}{3}; \frac{10\pi}{3};$

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

No solution exists

Explanation

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

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

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

$\cos 4x = \cos 2x$

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

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

$x=0; \frac{2\pi}{3}; \frac{4\pi}{3}; 2\pi; \frac{8\pi}{3}; \frac{10\pi}{3};$

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

No solution exists

Explanation

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

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

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

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

$x=0, \frac{7\pi}{3}, \frac{11\pi}{3}$

$x=0, \frac{7\pi}{6}, \frac{11\pi}{6}$

$x=0, \frac{\pi}{6}, \frac{5\pi}{6}$

$x=0, \frac{\pi}{3}, \frac{5\pi}{3}$

Explanation

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

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

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

$x=0, \frac{7\pi}{3}, \frac{11\pi}{3}$

$x=0, \frac{7\pi}{6}, \frac{11\pi}{6}$

$x=0, \frac{\pi}{6}, \frac{5\pi}{6}$

$x=0, \frac{\pi}{3}, \frac{5\pi}{3}$

Explanation

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

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

$4\cos ^2x=3$

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

$x=\frac{\pi}{6}; \frac{11\pi}{6}$

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

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

No solution exists

Explanation

$4\cos ^2x=3$ ; 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,