Question

Determine all solutions to the equation radical 2 times cosine 2 times x equals sine squared x plus cosine squared x on the interval [0, 2π).

Answer 1

Given

The equation,

$\sqrt{2}\cos2x=\sin^2x+\cos^2x$

To determine all the solutions in the interval [0, 2π).

Explanation:

It is given that,

$\sqrt{2}\cos2x=\sin^2x+\cos^2x$

Since

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

Then,

$\begin{gathered} \sqrt{2}\cos2x=\sin^2x+\cos^2x \\ \Rightarrow\cos2x=\frac{1}{\sqrt{2}} \\ \Rightarrow2x=\cos^{-1}(\frac{1}{\sqrt{2}}) \\ \Rightarrow2x=\frac{\pi}{4} \\ \Rightarrow x=\frac{\pi}{8} \end{gathered}$

Hence, the solutions of the given equation in [0, 2π) is,

$a)\text{ }x=\frac{\pi}{8},\frac{7\pi}{8},\frac{9\pi}{8},\frac{15\pi}{8}$