How do you solve $4sin^2x=2cosx+1$ and find all solutions in the interval $[0,2pi)$ ?

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Answer 1 · 2018-03-19T19:10:13

There are two solutions, $cos^-1((sqrt(13)-1)/4)$ , and $2pi-cos^-1((sqrt(13)-1)/4)$

Use the Pythagorean Theorem to replace $sin^2x$ with $1-cos^2x$ .

$4(1-cos^2x)=2cosx+1$

Use the distributive property

$4-4cos^2x=2cosx+1$

Put this "quadratic" equation in standard form.

$4cos^2x+2cosx-3=0$

Use the quadratic formula to solve for $cosx$ .

$cosx=(-2+-sqrt(2^2-4*4*3))/(2*4)=(-2+-sqrt(52))/(2*4)=(-1+-sqrt(13))/(4)$

We can throw out $(-1-sqrt(13))/4$ as a real root because it is less than -1, and $cosx$ cannot be less than -1 if $x$ is real.

Therefore we need only find angles between 0 and $2pi$ whose cosine are $(sqrt(13)-1)/4$ . These angle would be

$x=cos^-1((sqrt(13)-1)/4)$ and $x=2pi-cos^-1((sqrt(13)-1)/4)$ .

How do you solve 4sin^2x=2cosx+14sin^2x=2cosx+1 and find all solutions in the interval [0,2pi)[0,2pi)?