Question

Question Part Points Submissions Used Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.) cos(2θ) + sin^2(θ) = 0

Answer 1

$\bf \textit{Double Angle Identities} \\\\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ \boxed{1-2sin^2(\theta)}\\ 2cos^2(\theta)-1 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ cos(2\theta )+sin^2(\theta )=0\implies \boxed{1-2sin^2(\theta)}+sin^2(\theta )=0 \\\\\\ 1-sin^2(\theta )=0\implies 1=sin^2(\theta )\implies \pm\sqrt{1}=sin(\theta ) \\\\\\ \pm 1=sin(\theta )\implies sin^{-1}(\pm 1) = \theta \implies \theta = \begin{cases} \frac{\pi }{2}\\\\ \frac{3\pi }{2} \end{cases}$