Question

Trig. Solve, finding all solutions in [0,2pi).
6cosx+6sinx=3sqrt(6)

Answer 1

\[\cos x+\sin x=\frac{\sqrt{6}}{2}\]
\[\cos ^{2}x+2\sin x \cos x+\sin ^{2}x=\frac{6}{4}\]
\[1+2\sin x \cos x=\frac{3}{2}\]
\[2\sin x \cos x=\frac{1}{2}\]
\[\sin 2x=\frac{1}{2}\]

Answer 2

Perhaps you can take it from there.

Answer 3

\[2x=30, 150, 390, 510\]
\[x=15, 75, 195, 255\]

Answer 4

This is an interesting question. I don't know if my approach is "conventional," but I think it will work.\[6cosx+6sinx=3\sqrt(6)\implies cosx+sinx=\sqrt{6}/2\]\[\implies \cos^2x+\sin^2x+2cosxsinx=3/2\implies cosxsinx=1/4\]\[\implies \sin^2xcos^2x=1/16\implies (1-\cos^2x)\cos^2x=1/16\]\[\implies 0=\cos^4x-\cos^2x+16\]So let u=cos^2x and solve. It doesn't seem quite as straightforward as Mertsj's method, but should work.

Answer 5

I think my failure to remember those double angle formulas was a substantial drawback on this particular problem.

Answer 6

ok i'm still tryin to figure it out thanks

Answer 7

You don't understand what I posted?

Answer 8

i get it now thanks!!!