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!

How do you solve 2cos^2theta-sin^2theta=12cos^2theta-sin^2theta=1?