Question

solve ( 2 cos ( 2 \theta ) = 2 sin ^ { 2 } ( \theta ) + 1 ) for all solutions ( 0 leq \theta < 2 pi ).

( \theta = )

give your answers accurate to at least 2 decimal places and in a list separated by commas.

question help: video

Explanation:

Step1: Use double - angle formula

Recall the double - angle formula \(\cos(2\theta)=1 - 2\sin^{2}\theta\). Substitute it into the equation \(2\cos(2\theta)=2\sin^{2}\theta + 1\).
We get \(2(1 - 2\sin^{2}\theta)=2\sin^{2}\theta+1\).

Step2: Expand and simplify the equation

Expand the left - hand side: \(2-4\sin^{2}\theta=2\sin^{2}\theta + 1\).
Move all terms involving \(\sin^{2}\theta\) to one side: \(-4\sin^{2}\theta-2\sin^{2}\theta=1 - 2\).
Combine like terms: \(-6\sin^{2}\theta=-1\).
Then \(\sin^{2}\theta=\frac{1}{6}\), so \(\sin\theta=\pm\frac{1}{\sqrt{6}}\approx\pm0.41\).

Step3: Solve for \(\theta\)

When \(\sin\theta=\frac{1}{\sqrt{6}}\), \(\theta=\sin^{-1}(\frac{1}{\sqrt{6}})\approx0.43\) or \(\theta=\pi-\sin^{-1}(\frac{1}{\sqrt{6}})\approx2.71\).
When \(\sin\theta=-\frac{1}{\sqrt{6}}\), \(\theta=\pi+\sin^{-1}(\frac{1}{\sqrt{6}})\approx3.57\) or \(\theta = 2\pi-\sin^{-1}(\frac{1}{\sqrt{6}})\approx5.85\).

Answer:

\(0.43,2.71,3.57,5.85\)