Question

Solve the following equation exactly on the interval 0 ≤ θ ≤ 2π ? cos 2θ + sin θ = 0

Answer 1

`\bf \textit{Double Angle Identities} \\ \quad \\ sin(2\theta)=2sin(\theta)cos(\theta) \\ \quad \\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ \boxed{1-2sin^2(\theta)}\\ 2cos^2(\theta)-1 \end{cases} \\ \quad \\\\ tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\ -------------------------------`


`\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} (7\pi )/(6)\\\\ (11\pi )/(6) \end{cases}\\\\ -------------------------------\\\\ 0=sin(\theta )-1\implies 1=sin(\theta )\implies \measuredangle \theta =(\pi )/(2)`