# trigonomentry

Solve for θ from 0 to 360 degrees [sin(2θ)]²- (cosθ+sinθ)=

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asked by pooo
1. looks like your equation to solve is incomplete.

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posted by Reiny
2. =1

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posted by pooo
3. you can rewrite it as

sinx + cosx + cos^2(2x) = 0

clearly x= 3π/4, π, 7π/4 are some solutions.
To find them all will take some doing.

working with our equation,
-sinx = cos^2(2x) + cosx
square both sides to get

sin^2(x) = cos^4(2x) - 2cos^2(2x)cos(x) + cos^2(x)

1-cos^2(x) = (2cos^2(x)-1)^4 - 2cos(x)(2cos^2(x)-1)^2 + cos^2(x)

Or, letting u = cosx,
1-u^2 = (2u^2-1)^4-2u(2u^2-1)^2 + u^2

16u^8-32u^6-8u^5+24u^4+8u^3-6u^2-2u = 0

2u(8u^7-16u^5-4u^4+12u^3+4u^2-3u+1) = 0

Not sure how good you are at solving such polynomials. There will also be some extraneous solutions from the squaring.

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posted by Steve
4. oh, yes. x=3π/2 also works.

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posted by Steve

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