Question

solve for the angle θ where 0 ≤ θ ≤ 2π.
sin 2θ - √2 cos θ = 0

θ =
(use a comma to separate answers as needed. type an exact answer in terms of π.)

Explanation:

Step1: Use double - angle formula for sine

Recall the double - angle formula \(\sin2\theta = 2\sin\theta\cos\theta\). Substitute this into the given equation \(\sin2\theta-\sqrt{2}\cos\theta = 0\).
We get \(2\sin\theta\cos\theta-\sqrt{2}\cos\theta = 0\).

Step2: Factor out \(\cos\theta\)

Factor \(\cos\theta\) from the left - hand side of the equation: \(\cos\theta(2\sin\theta - \sqrt{2})=0\).

Step3: Set each factor equal to zero

We have two cases:

Case 1: \(\cos\theta = 0\)

For the equation \(\cos\theta = 0\) and \(0\leq\theta\leq2\pi\), we know that \(\cos\theta = 0\) when \(\theta=\frac{\pi}{2}\) or \(\theta=\frac{3\pi}{2}\) (since the cosine function is zero at odd multiples of \(\frac{\pi}{2}\) in the interval \([0,2\pi]\)).

Case 2: \(2\sin\theta-\sqrt{2}=0\)

First, solve the equation \(2\sin\theta-\sqrt{2}=0\) for \(\sin\theta\).
Add \(\sqrt{2}\) to both sides: \(2\sin\theta=\sqrt{2}\).
Then divide both sides by 2: \(\sin\theta=\frac{\sqrt{2}}{2}\).
For the equation \(\sin\theta=\frac{\sqrt{2}}{2}\) and \(0\leq\theta\leq2\pi\), we know that \(\sin\theta=\frac{\sqrt{2}}{2}\) when \(\theta = \frac{\pi}{4}\) or \(\theta=\frac{3\pi}{4}\) (since the sine function has a value of \(\frac{\sqrt{2}}{2}\) in the first and second quadrants).

Answer:

\(\frac{\pi}{4},\frac{3\pi}{4},\frac{\pi}{2},\frac{3\pi}{2}\)