If 2^(2sin^2 θ−3sin θ+1) + 2^(2−2sin^2 θ+3sin θ) = 9 find the value of sin θ+cos θ .


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Question Number 137851 by liberty last updated on 07/Apr/21

$I f 2^{2 \sin^{2} θ - 3 \sin θ + 1} + 2^{2 - 2 \sin^{2} θ + 3 \sin θ} = 9$ $f i n d t h e v a l u e o f \sin θ + \cos θ .$
	Answered by bobhans last updated on 07/Apr/21
	$consider 2^(2−2sin^2 θ+3sin θ)) = 4.2^(−2sin^2 θ+3sin θ) and 2^(2sin^2 θ−3sin θ+1) = 2.2^(2sin^2 θ−3sin θ) let 2^(2sin^2 θ−3sin θ ) = b we get (4/b) + 2b = 9 ; 2b^2 −9b+4 = 0 (2b−1)(b−4) = 0 { ((b=(1/2))),((b=4)) :} case(1) b=(1/2) ⇒2^(2sin^2 θ−3sin θ) = 2^(−1) we get 2sin^2 θ−3sin θ+1=0 (2sin θ−1)(sin θ−1)=0 { ((sin θ=(1/2)∧cos θ=±((√3)/2) )),((sin θ=1 ∧cos θ=0 (no solution))) :} case(2) b=4⇒2^(2sin^2 θ−3sin θ) =2^2 we get 2sin^2 θ−3sin θ−2=0 (2sin θ−1)(sin θ+2)=0 →sin θ=(1/2) same to case(1) then sin θ+cos θ = ((1±(√3))/2)$
	$c o n s i d e r 2^{2 - 2 \sin^{2} θ + 3 \sin θ)} =$ $4 . 2^{- 2 \sin^{2} θ + 3 \sin θ} a n d 2^{2 \sin^{2} θ - 3 \sin θ + 1} =$ $2 . 2^{2 \sin^{2} θ - 3 \sin θ}$ $l e t 2^{2 \sin^{2} θ - 3 \sin θ} = b w e g e t$ $\frac{4}{b} + 2 b = 9; 2 b^{2} - 9 b + 4 = 0$ $(2 b - 1) (b - 4) = 0 {\begin{cases} b = \frac{1}{2} \\ b = 4 \end{cases}$ $c a s e (1) b = \frac{1}{2} \Rightarrow 2^{2 \sin^{2} θ - 3 \sin θ} = 2^{- 1}$ $w e g e t 2 \sin^{2} θ - 3 \sin θ + 1 = 0$ $(2 \sin θ - 1) (\sin θ - 1) = 0$ ${\begin{cases} \sin θ = \frac{1}{2} \land \cos θ = \pm \frac{\sqrt{3}}{2} \\ \sin θ = 1 \land \cos θ = 0 (n o s o l u t i o n) \end{cases}$ $c a s e (2) b = 4 \Rightarrow 2^{2 \sin^{2} θ - 3 \sin θ} = 2^{2}$ $w e g e t 2 \sin^{2} θ - 3 \sin θ - 2 = 0$ $(2 \sin θ - 1) (\sin θ + 2) = 0$ $\to \sin θ = \frac{1}{2} s a m e t o c a s e (1)$ $t h e n \sin θ + \cos θ = \frac{1 \pm \sqrt{3}}{2}$
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