How do you solve $sin^2 x = 2cosx + 2$ from 0 to 2pi?

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Answer 1 · 2018-05-07T02:37:03

$color(purple)(=> pi$

$sin^2x = 2 cos x + 2$

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

$cos^2 x + 2 cos x + 2 - 1 = 0$

$cos^2 x + 2 cos x + 1 = 0$

$(cos x = 1)^2 = 0$

$cos x + 1 = 0$

$cos x = -1$

$color(purple)(x = cos ^-1 -1 = cos^-1 cos (pi) = pi$

How do you solve sin^2 x = 2cosx + 2sin^2 x = 2cosx + 2 from 0 to 2pi?