Attention Required!
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
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asked May 28, 2019 at 14:09
Diya SarkarDiya Sarkar
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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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