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Leah Wilkinson
Mathematics
30 September, 03:21
2cos^2x+cosx-1=0
Find the degree solutions
+5
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Devyn Graves
30 September, 03:32
0
2cos²x+cos x-1=0
cos x=t
2t²+t-1=0
t=[-1⁺₋√ (-1+8) ]/2 = (-1⁺₋3) / 4
We have two possible set solutions:
First set solutions.
t₁ = (-1-3) / 2=-4/4=-1
cos x=-1 ⇒x=cos⁻¹ (-1) = π + 2kπ or 180º+360ºk (k = ( ... - 2,-1,0,1,2 ...)
Second set solutions:
t₂ = (-1+3) / 4=2/4=1/2
cos x=1/2 ⇒ x=cos⁻¹ 1/2=π/3+2kπ U 5π/3 + 2kπ or
60º+360ºK U 300º+360ºK (k = ... - 2,-1,0,1,2, ...)
solutions: first set solutions U second set solutions:
Answer in radians : π + 2kπ U π/3+2kπ U 5π/3 + 2kπ (k = ... - 1,0,1, ...)
Answer is degrees: 180º+360ºk U 60º+360ºK U 300º+360ºK (k = ... - 2,-1,0,1,2, ...)
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