Question

Use trigonometric identities and algebraic methods, as necessary, to solve the following trigonometric equation. Please identify all possible solutions by including all answers in [0,2Ï€) and indicating the remaining answers by using n to represent any integer. Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution." cos2(5x)=sin2(5x)

Answer 1

The trigonometric equation 2cos^2(x) + 2 = 4 has solutions x = 0 and x = π in the interval [0, 2π). To include all possible solutions, we express them as x = 2nπ and x = π + 2nπ with n as any integer.

To solve the trigonometric equation 2cos^2(x) + 2 = 4, we start by simplifying the equation:

1. Subtract 2 from both sides to get 2cos^2(x) = 2.
2. Divide both sides by 2 to get cos^2(x) = 1.
3. Take the square root of both sides, considering both positive and negative roots, to get cos(x) = ±1.
4. Find angles x in the interval [0, 2π) where the cosine is ±1. For cos(x) = 1, x = 0. For cos(x) = -1, x = π.

This equation has two solutions in the interval [0, 2π): x = 0 and x = π (which are 0 and 3.1416 when rounded to four decimal places). To express the infinite set of solutions that occur at every period of cos(x), we add 2nπ to each solution, where n is any integer, giving the final solutions as x = 2nπ and x = π + 2nπ.

the complete Question is given below:

Use trigonometric identities and algebraic methods, as necessary, to solve the following trigonometric equation. Please identify all possible solutions by including all answers in [0, 2π)

and indicating the remaining answers by using n to represent any integer. Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution."

2cos^2 (x) + 2 = 4

Enter your answer in radians, as an exact answer when possible. Multiple answers should be separated by commas.

Answer 2

`\large\boxed{x=\pm(3\pi)/(20)+(2n\pi)/(5)\ \vee\ x=\pm(\pi)/(20)+(2n\pi)/(5)}`

Explanation:

`[tex]\cos^2(5x)=\sin^2(5x)\qquad\text{substitute}\ t=5x\\\\\cos^2t=\sin^2t\qquad\text{use}\ \sin^2\theta+\cos^2\theta=1\to\sin^2\theta=1-\cos^2\theta\\\\\cos^2t=1-\cos^2t\qquad\text{add}\ \cos^2t\ \text{to both sides}\\\\2\cos^2t=1\qquad\text{divide both sides by 2}\\\\\cos^2t=(1)/(2)\Rightarrow \cos t=\pm\sqrt{(1)/(2)}\\\\\cos t=\pm(\sqrt2)/(2)\\\\\cos t=-(\sqrt2)/(2)\Rightarrow t=\pm(3\pi)/(4)+2n\pi\\\\\cos t=(\sqrt2)/(2)\Rightarrow t=\pm(\pi)/(4)+2n\pi`

`t=5x\\\\5x=\pm(3\pi)/(4)+2n\pi\ \vee\ 5x=\pm(\pi)/(4)+2n\pi\qquad\text{divide both sides by 5}\\\\x=\pm(3\pi)/(20)+(2n\pi)/(5)\ \vee\ x=\pm(\pi)/(20)+(2n\pi)/(5)`