Mathematics, 28.06.2019 03:50 studyowl1087
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
Answers: 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}](/tpl/images/0025/7702/b5bfe.png)
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Question part points submissions used use a double- or half-angle formula to solve the equation in t...
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