Solve: 3sin⁡2x+4cos⁡2x−2cos⁡x+6sin⁡x−6=0 My Try 6sin⁡xcos⁡x+4(cos2x−sin2x)−2cos⁡x+6sin⁡x−6=0

Question

Solve: 3sin⁡2x+4cos⁡2x−2cos⁡x+6sin⁡x−6=0  My Try   6sin⁡xcos⁡x+4(cos2x−sin2x)−2cos⁡x+6sin⁡x−6=0

Timothy Wolff · Accepted Answer

6sin⁡xcos⁡x+4(cos2x−sin2x)−2cos⁡x+6sin⁡x−6=0   6sin⁡xcos⁡x−9sin2x−cos2x−2cos⁡x+6sin⁡x−1=0   −(3sin⁡x−cos⁡x)2+2(3sin⁡x−cos⁡x)−1=0  (3sin⁡x−cos⁡x)2−2(3sin⁡x−cos⁡x)+1=0

Philip Williams · Accepted Answer

x=2y  we get after some manipulation   −4cos4y−36cos2ysin2y+24cos3ysin⁡y=0   −4cos2y(cos⁡y−3sin⁡y)2=0

Vasquez · Accepted Answer

Abbreviating sin⁡x and cos⁡x to s and c, and writing sin⁡2x=2sc and cos⁡2x=2c2−1, the equation becomes 6sc+4(2c2−1)−2c+6s−6=0 Factoring out a 2 and grouping terms, we rewrite this as 3s(c+1)+4c2−c−5=0 But 4c2−c−5=(4c−5)(c+1), so we have two solutions in s and c: c=-1 and 3s+4c=5 Reverting to sin⁡x and cos⁡x and recognizing the Pythagorean triple 32+42=52, these become cos⁡x=−1 and sin⁡(x+θ)=1 where sin⁡θ=45 and cos⁡θ=35, e.g. θ=arcsin⁡(45). So the solution set is {π+2nπ|n∈Z}∪{π2−arcsin⁡(45)+2nπ|n∈Z}

Solve: 3\sin 2x+4 \cos 2x− 2\cos x+6\sin x-6=0 My Try 6 \sin

Answered question

Answer & Explanation

New Questions in Precalculus