Questions: Solve the equation in radians for all exact solutions where appropriate. Write answers using least possible nonnegative angle measures. -√2 cos x - 1 = cos 2x

Solution Steps

We start with the equation: \[ -\sqrt{2} \cos x - 1 = \cos 2x \] Using the double angle identity for cosine, we rewrite \(\cos 2x\) as: \[ \cos 2x = 2\cos^2 x - 1 \] Thus, the equation becomes: \[ -\sqrt{2} \cos x - 1 = 2\cos^2 x - 1 \]

Rearranging the equation gives us: \[ -\sqrt{2} \cos x = 2\cos^2 x \] This can be rewritten as: \[ 2\cos^2 x + \sqrt{2} \cos x = 0 \]

Factoring out \(\cos x\) from the equation, we have: \[ \cos x (2\cos x + \sqrt{2}) = 0 \] This gives us two cases to solve:

1. \(\cos x = 0\)
2. \(2\cos x + \sqrt{2} = 0\)

Case 1: For \(\cos x = 0\): \[ x = \frac{\pi}{2} + k\pi \quad (k \in \mathbb{Z}) \] The solutions within the range \([0, 2\pi)\) are: \[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \]

Case 2: For \(2\cos x + \sqrt{2} = 0\): \[ \cos x = -\frac{\sqrt{2}}{2} \] The solutions for this case are: \[ x = \frac{3\pi}{4} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{4} + 2k\pi \quad (k \in \mathbb{Z}) \] The solutions within the range \([0, 2\pi)\) are: \[ x = \frac{3\pi}{4}, \quad x = \frac{5\pi}{4} \]

Combining all the solutions from both cases, we have: \[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{4}, \quad x = \frac{5\pi}{4}, \quad x = \frac{3\pi}{2} \]

Final Answer