Mathematics, 17.07.2019 17:50 jlang8
Solve the following equation exactly on the interval 0 ≤ θ ≤ 2π ? cos 2θ + sin θ = 0
Answers: 1
![\bf cos(2\theta )+sin(\theta )=0\implies 1-2sin^2(\theta )+sin(\theta )=0 \\\\\\ 0=2sin^2(\theta )-sin(\theta )-1\implies 0=[2sin(\theta )+1][sin(\theta )-1]\\\\ -------------------------------\\\\ 0=2sin(\theta )+1\implies -1=2sin(\theta )\implies -\cfrac{1}{2}=sin(\theta ) \\\\\\ \measuredangle \theta = \begin{cases} \frac{7\pi }{6}\\\\ \frac{11\pi }{6} \end{cases}\\\\ -------------------------------\\\\ 0=sin(\theta )-1\implies 1=sin(\theta )\implies \measuredangle \theta =\frac{\pi }{2}](/tpl/images/0101/0032/9f995.png)
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Solve the following equation exactly on the interval 0 ≤ θ ≤ 2π ? cos 2θ + sin θ = 0...
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