# Attention Required!

Find the number of solutions to lớn \$sin x+sin2x+sin3x=cos x+cos2x+cos3x\$ with \$xin<0,2pi)\$

I tried lớn solve it by doing this:-

\$\$2sin(2x)cos x + sin(2x) = 2cos(2x)cos x + cos(2x)\$\$

\$\$sin(2x) (2cos x+1) = cos(2x) (2cos x + 1)\$\$

\$\$ sin(2x) = cos(2x)\$\$

\$\$ 2cos xsin x = 1 - 2sin^2 x\$\$

\$\$ 2cos xsin x + 2sin^2 x - 1 = 0\$\$

\$\$ 2sin x(cos x + sin x) -1=0\$\$

\$\$ 2sin x(2sin x) -1=0\$\$

\$\$ 4sin^2 x = 0\$\$

\$\$ sin^2 x = (frac12)^2\$\$

\$\$ sin^2 x = sin^2 (fracpi6)\$\$

Now by using \$x = nπ -fracpi6\$ I got 4 solutions between \$0,2pi\$. But the answer is 6. Where I did wrong. Please don"t mind for mistakes và answer ASAP. THANKS IN ADVANCE.

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proof-verification trigonometry
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edited May 28, 2019 at 14:44 Michael Rozenberg
asked May 28, 2019 at 14:09 Diya SarkarDiya Sarkar
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6
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You have also \$\$2cosx+1=0,\$\$ which gives another two roots.

Because \$\$sin2x(2cosx+1)=cos2x(2cosx+1)\$\$ it"s\$\$sin2x(2cosx+1)-cos2x(2cosx+1)=0\$\$ or\$\$(2cosx+1)(sin2x-cos2x)=0\$\$ & we obtain:\$\$2cosx+1=0\$\$ or\$\$sin2x-cos2x=0.\$\$

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edited May 28, 2019 at 14:23
answered May 28, 2019 at 14:17 Michael RozenbergMichael Rozenberg
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