How do you solve #2cos^2theta-sin^2theta=1#?

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Answer 1 · 2016-12-31T21:33:35

Use the identity #sin^2x + cos^2x = 1#.

#2cos^2theta - (1 - cos^2theta) = 1#

#:.2cos^2theta - 1 + cos^2theta = 1#

#:.3cos^2theta = 2#

#:.cos^2theta = 2/3#

#:.costheta = +-sqrt(2/3)#

#:.theta = arccossqrt(2/3) + 2pin, pi - arccossqrt(2/3) + 2pin, pi + arccossqrt(2/3) + 2pin, 2pi - arccossqrt(2/3) + 2pin#

#:.theta = arccossqrt(2/3) + 2pin, (2n + 1)pi - arccossqrt(2/3), (2n + 1)pi + arccossqrt(2/3), 2pi(1 + n) - arccossqrt(2/3)#

Hopefully this helps!