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Question Number 15263 by Tinkutara last updated on 08/Jun/17

$The equation a \sin x + \cos 2 x = 2 a - 7$ $possesses a solution if$ $(1) a > 6$ $(2) 2 ⩽ a ⩽ 6$ $(3) a > 2$ $(4) a$
	Answered by ajfour last updated on 08/Jun/17

	$a \sin x + 1 - 2 \sin^{2} x = 2 a - 7$ $\sin^{2} x - \frac{a}{2} \sin x = 4 - a$ ${(\sin x - \frac{a}{4})}^{2} = \frac{a^{2}}{16} - a + 4$ ${(\sin x - \frac{a}{4})}^{2} = {(\frac{a}{4} - 2)}^{2}$ $∣ \sin x - \frac{a}{4} ∣=∣ \frac{a}{4} - 2 ∣$ $\Rightarrow \frac{a}{4} - 2 = \pm ∣ \sin x - \frac{a}{4} ∣$ $w i t h t h e - v e s i g n, a b o v e e q n . \Rightarrow$ $n o v a l u e o f a t o f u l f i l l t h e c o n d i t i o n;$ $w i t h t h e + v e s i g n i t i m p l i e s$ $\frac{a}{2} = 2 + \sin x$ $o r a = 4 + 2 \sin x$ $A s - 1 ⩽ \sin x ⩽ 1$ $w e c a n c o n c l u d e t h a t$ $2 ⩽ a ⩽ 6 .$
	Commented by Tinkutara last updated on 09/Jun/17

	$Thanks Sir!$
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