How do you solve the identity $2 sin^2 x = 2 + cos x$ ?

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Answer 1 · 2015-08-06T13:34:33

$x = (2n-1)\frac{\pi}{2} " or " x = (2n-1)\pi \pm {\pi}/3 , " " n inZZ$

$sin^2 x + cos^2 x -= 1$
$\therefore 2sin^2 x = 2-2cos^2 x$

Substituting above to original equation:
$2-2cos^2 x = 2 + cos x$

Re-arrange:
$2cos^2 x + cos x = 0$
$cos x(2cos x + 1) = 0$

$cos x = 0 " or " cos x = -\frac{1}{2}$

$x = (2n-1)\frac{\pi}{2} " or " x = (2n-1)\pi \pm {\pi}/3 , " " n in ZZ$

How do you solve the identity 2 sin^2 x = 2 + cos x 2 sin^2 x = 2 + cos x?