Solve the equation for solutions in the interval [0°, 360°). Round to the nearest degree.
-$\sin 2 \theta + \cos 2 \theta = 1$

Question

Quizplus · Accepted Answer

To solve $\sin 2\theta + \cos 2\theta = 1$, use the identity $\cos^2 2\theta + \sin^2 2\theta = 1$. Subtract $\sin 2\theta$ from both sides to get $\cos^2 2\theta - \sin 2\theta = 0$. Let $x = \sin 2\theta$, then $(1-x^2) - x = 0$, solving this gives $x = 0$ or $x = \sqrt{2}/2$. Thus, $2\theta = 0^\circ, 45^\circ, 180^\circ, 225^\circ$, leading to $\theta = 0^\circ, 22.5^\circ, 90^\circ, 112.5^\circ$. Doubling these angles to revert to the original variable gives the solutions $0^\circ, 45^\circ, 180^\circ, 225^\circ$.

Solve the Equation for Solutions in the Interval [0°, 360°) sin⁡2θ+cos⁡2θ=1\sin 2 \theta + \cos 2 \theta = 1sin2θ+cos2θ=1

Solve the Equation for Solutions in the Interval [0°, 360°) $\sin 2 \theta + \cos 2 \theta = 1$