Solve... 1+ sin 2x = sin x + cos x


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Question Number 1961 by alib last updated on 26/Oct/15

$S o l v e . . .$ $1 + s i n 2 x = s i n x + c o s x$
	Answered by Rasheed Soomro last updated on 26/Oct/15

	$1 + s i n 2 x = s i n x + c o s x$ $1 + 2 s i n x c o s x - s i n x - c o s x = 0$ ${s i n}^{2} x + {c o s}^{2} x + 2 s i n x c o s x - (s i n x + c o s x) = 0$ ${(s i n x + c o s x)}^{2} - (s i n x + c o s x) = 0$ $(s i n x + c o s x) (s i n x + c o s x - 1) = 0$ $s i n x + c o s x = 0 ∣ s i n x + c o s x - 1 = 0$ $s i n x + c o s x = 0 ∣ s i n x + c o s x = 1$ $s i n x = - c o s x ∣ s i n x = 1 - c o s x$ ${s i n}^{2} x = {c o s}^{2} x ∣ {s i n}^{2} x = 1 + {c o s}^{2} x - 2 c o s x$ ${s i n}^{2} x = 1 - {s i n}^{2} x ∣ 1 - {c o s}^{2} x = 1 + {c o s}^{2} x - 2 c o s x$ ${s i n}^{2} x = \frac{1}{2} ∣ {c o s}^{2} x - c o s x = 0$ $x = {s i n}^{- 1} (\pm \frac{1}{\sqrt{2}}) ∣ c o s x (c o s x - 1) = 0 \Rightarrow c o s x = 0 \lor c o s x = 1$ $x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4} i n t h e i n t e r v a l (0, 2 π) ∣ x = \frac{π}{2}, \frac{3 π}{2}, 0 i n (0, 2 π)$ $E x t r a n e o u s r o o t s m a y b e i n l u d e d .$
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