Solve the equation 32sin⁡(x)−cos⁡x=cos2x My approach cos2x=1−sin2x 32sin⁡x−cos2x=cos⁡x 32sin⁡x+sin2x−1=cos⁡x 3sin⁡x+2sin2x−2=2cos⁡x Though the equation comes in form of sin⁡x from here onward after squaring still not getting the answer.

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Solve the equation 32sin⁡(x)−cos⁡x=cos2x  My approach cos2x=1−sin2x  32sin⁡x−cos2x=cos⁡x   32sin⁡x+sin2x−1=cos⁡x   3sin⁡x+2sin2x−2=2cos⁡x   Though the equation comes in form of sin⁡x from here onward after squaring still not getting the answer.

trnovitom06 · Accepted Answer

A straightforward approach is indeed to square the equation:  32sin⁡x=cos2x+cos⁡x  and replace sin2x  by  1−cos2x. You obtain a quartic equation in cosx. Maybe not the most elegant solution, but you cannot go wrong with it. Also, one root is clear without calculation: cos⁡x=12 is fine. So you can surely reduce to a cubic equation. (And possibly even further to a quadratic one.)

vasselefa · Accepted Answer

Just to flesh out first answer, let c=cos⁡x, s=sin⁡x so s32=c(1+c) and 3(1−c2)=4c2(1+c)2. After some rearrangment, (c+1)(c−12)(2c2+3c+3)=0 , with the quadratic factor lacking real roots. We must be careful with the signs of c,s. One solution is c=−1,s=0, other is c=12, s=32. In other words, the real x allowed are x=π(2k+1),,x=π(1+6k)3  for  k∈Z

Solve the equation \frac{\sqrt {3}}{2}\sin(x) -\cos x=\cos^2x My approach \cos^2x=1-\sin^2x \frac{\sqrt {3}}{2}\sin

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