anonymous
  • anonymous
Trig. Solve, finding all solutions in [0,2pi). 6cosx+6sinx=3sqrt(6)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Mertsj
  • Mertsj
\[\cos x+\sin x=\frac{\sqrt{6}}{2}\] \[\cos ^{2}x+2\sin x \cos x+\sin ^{2}x=\frac{6}{4}\] \[1+2\sin x \cos x=\frac{3}{2}\] \[2\sin x \cos x=\frac{1}{2}\] \[\sin 2x=\frac{1}{2}\]
Mertsj
  • Mertsj
Perhaps you can take it from there.
Mertsj
  • Mertsj
\[2x=30, 150, 390, 510\] \[x=15, 75, 195, 255\]

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anonymous
  • anonymous
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.
anonymous
  • anonymous
I think my failure to remember those double angle formulas was a substantial drawback on this particular problem.
anonymous
  • anonymous
ok i'm still tryin to figure it out thanks
Mertsj
  • Mertsj
You don't understand what I posted?
anonymous
  • anonymous
i get it now thanks!!!

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