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#