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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