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

Jason Yuhas

Jason Yuhas

Answered question

2021-12-31

Solve: 3sin2x+4cos2x2cosx+6sinx6=0
My Try
6sinxcosx+4(cos2xsin2x)2cosx+6sinx6=0

Answer & Explanation

Timothy Wolff

Timothy Wolff

Beginner2022-01-01Added 26 answers

6sinxcosx+4(cos2xsin2x)2cosx+6sinx6=0
6sinxcosx9sin2xcos2x2cosx+6sinx1=0
(3sinxcosx)2+2(3sinxcosx)1=0
(3sinxcosx)22(3sinxcosx)+1=0
Philip Williams

Philip Williams

Beginner2022-01-02Added 39 answers

x=2y we get after some manipulation
4cos4y36cos2ysin2y+24cos3ysiny=0
4cos2y(cosy3siny)2=0
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

Abbreviating sinx and cosx to s and c, and writing sin2x=2sc and cos2x=2c21, the equation becomes
6sc+4(2c21)2c+6s6=0
Factoring out a 2 and grouping terms, we rewrite this as
3s(c+1)+4c2c5=0
But 4c2c5=(4c5)(c+1), so we have two solutions in s and c:
c=-1 and 3s+4c=5
Reverting to sinx and cosx and recognizing the Pythagorean triple 32+42=52, these become
cosx=1 and sin(x+θ)=1
where sinθ=45 and cosθ=35, e.g. θ=arcsin(45). So the solution set is
{π+2nπ|nZ}{π2arcsin(45)+2nπ|nZ}

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