Solve the following for 0 ≤ x ≥2¶
2sin²(x)-cos(x)=1

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Originally Posted by mkquadri

Solve the following for 0 ≤ x ≥2¶
2sin²(x)-cos(x)=1

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You need to use one of the laws to break it down. I've been out of calculus for a year now, but I believe that you can find them almost anywhere online. There's something that takes sin and changes it to a sin and cos so that you can make it easier to work with.

Quote:

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Originally Posted by mkquadri

Solve the following for 0 ≤ x ≥2¶
2sin²(x)-cos(x)=1

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2(1 - cos^2 x) - cos x = 1 [apply the 1st Pythagorean Identity]
2 - 2cos^2 x - cos x = 1 [distribute]
-2cos^2 x - cos x + 1 = 0 [set = 0 and put in descending powers]
2cos^2 x + cos x - 1 = 0 [multiply by -1 to eliminate the leading negative]
(2cos x - 1)(cos x + 1) = 0 [factor]
2cos x - 1 = 0 cos x + 1 = 0
2cos x = 1 cos x = -1
cos x = 1/2 cos x = -1

x = pi/3 , 5pi/3 , pi