Questions: Solve the following equation on the interval (0 leq theta<2 pi). (-cos ^2 theta+sin ^2 theta=1+cos theta)

Solution Steps

To solve the given trigonometric equation on the interval \(0 \leq \theta < 2\pi\), we can use trigonometric identities and algebraic manipulation. First, recall that \(\sin^2 \theta + \cos^2 \theta = 1\). Using this identity, we can express \(\sin^2 \theta\) in terms of \(\cos^2 \theta\) and simplify the equation. Then, we solve for \(\theta\) within the specified interval.

We start with the equation: \[ -\cos^2 \theta + \sin^2 \theta = 1 + \cos \theta \] Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\), we can rewrite the equation as: \[ -\cos^2 \theta + (1 - \cos^2 \theta) = 1 + \cos \theta \] This simplifies to: \[ -2\cos^2 \theta + 1 = 1 + \cos \theta \]

Rearranging gives us: \[ -2\cos^2 \theta - \cos \theta = 0 \] Factoring out \(-\cos \theta\) yields: \[ -\cos \theta (2\cos \theta + 1) = 0 \]

Setting each factor to zero gives us two cases:

1. \(-\cos \theta = 0\) which leads to \(\cos \theta = 0\)
2. \(2\cos \theta + 1 = 0\) which leads to \(\cos \theta = -\frac{1}{2}\)

For \(\cos \theta = 0\), the solutions in the interval \(0 \leq \theta < 2\pi\) are: \[ \theta = \frac{\pi}{2}, \frac{3\pi}{2} \]

For \(\cos \theta = -\frac{1}{2}\), the solutions are: \[ \theta = \frac{2\pi}{3}, \frac{4\pi}{3} \]

Combining all solutions, we have: \[ \theta = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{\pi}{2}, \frac{3\pi}{2} \]

Final Answer