Question

Solve over the interval [0, 2pi]

2sin^2x=cosx+1

Answer 1

`\textit{Pythagorean Identities} \\\\ \sin^2(\theta)+\cos^2(\theta)=1\implies \sin^2(\theta)=1-\cos^2(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}`

`2\sin^2(x)=\cos(x)+1\implies 2( ~~ 1-\cos^2(x) ~~ )=\cos(x)+1 \\\\\\ 2-2\cos^2(x)=\cos(x)+1\implies 0=2\cos^2(x)+\cos(x)-1 \\\\\\ 0=\stackrel{ \textit{notice, it's just a quadratic} }{2( ~~ \cos(x) ~~ )^2+\cos(x)-1}\hspace{5em}\stackrel{\textit{let's just for a second call}}{\cos(x)=C} \\\\\\ 0=2C^2+C-1\implies 0=(2C-1)(C+1) \\\\[-0.35em] ~\dotfill`

`0=2C-1\implies 1=2C\implies \cfrac{1}{2}=C\implies \cfrac{1}{2}=\cos(x) \\\\\\ \cos^(-1)\left( \cfrac{1}{2} \right)=x\implies {\Large \begin{array}{llll} x= \begin{cases} (\pi )/(3)\\\\ (5\pi )/(3) \end{cases} \end{array}} \\\\[-0.35em] ~\dotfill\\\\ 0=C+1\implies -1=C\implies -1=\cos(x) \\\\\\ \cos^(-1)(-1)=x\implies {\Large \begin{array}{llll} x=\pi \end{array}}`