Solve for θ from 0 to 360 degrees [sin(2θ)]²- (cosθ+sinθ)=
3 answers
looks like your equation to solve is incomplete.
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.
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.
oh, yes. x=3π/2 also works.