how do I find the exact solution for theta in sin2(θ)+2cos(θ)=−2?
Find the exact solution for theta in degrees
sin^2(θ)+2cos(θ)=−2, 0 degrees ≤ θ < 360 degrees
how do I solve this problem?
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Find the exact solution for theta in degrees
sin^2(θ)+2cos(θ)=−2, 0 degrees ≤ θ < 360 degrees
how do I solve this problem?
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Answers & Comments
well, sin^2(x) = 1 - cos^2(x)...
so
sin^2(x) + 2cos(x) = -2
==> 1 - cos^2(x) + 2cos(x) + 2 = 0
==> cos^2(x) - 2cos(x) - 3 = 0
==> (cos(x) -3)(cos(x) + 1) = 0
==> cos(x) = 3 (extraneous answer)
or
cos(x) = -1
==> x = 180 degrees
1-cos^2(θ)+2cos(θ)=−2
cos^2(θ) - 2cos(θ)-3 = 0
(cos(θ)-3)(cos(θ)+1) = 0
cos(θ) = -1
θ = 180