How do you solve $cos^2theta+2sintheta=-1$ ?

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Answer 1 · 2016-11-11T21:29:14

Please see the explanation.

Substitute $1 - sin^2(theta)$ for $cos^2(theta)$ :

$1 - sin^2(theta) + 2sin(theta) + 1 = 0$

Multiply by -1:

$sin^2(theta) - 2sin(theta) - 2 = 0$

Use the quadratic formula:

$sin(theta) = {2 +-sqrt((-2)^2 - 4(1)(-2))}/(2(1))$

$sin(theta) = 1 +-sqrt(3)$

Be must drop the + because it exceeds the domain of the sine function:

$sin(theta) = 1 -sqrt(3)$

$theta = sin^-1(1 -sqrt(3)) + 2pi$ and $theta = pi - sin^-1(1 -sqrt(3))$

If you wish, you may write these to repeat at integer multiples of $2pi$

How do you solve cos^2theta+2sintheta=-1cos^2theta+2sintheta=-1?