Question

Please help me with trigonometry. Use trigonometric identities to solve 2cos^2(θ)=sin(θ)+1 exactly for 0≤θ<2π. If there is more than one answer, enter your answers as a comma separated list.

Answer 1

`\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\ -------------------------------\\\\ 2cos^2(\theta )=sin(\theta )+1\implies 2[1-sin^2(\theta)]=sin(\theta )+1 \\\\\\ 2-2sin^2(\theta )=sin(\theta )+1\implies 0=2sin^2(\theta )+sin(\theta )-1 \\\\\\\ 0=[2sin(\theta )-1][sin(\theta )+1]\\\\ -------------------------------`


`\bf 0=2sin(\theta )-1\implies 1=2sin(\theta )\implies \cfrac{1}{2}=sin(\theta )\implies \measuredangle \theta = \begin{cases} (\pi )/(6)\\\\ (5\pi )/(6) \end{cases}\\\\ -------------------------------\\\\ 0=sin(\theta )+1\implies -1=sin(\theta )\implies \measuredangle \theta =(3\pi )/(2)`