Use the substitution t = sin(θ) to solve the equation 2sin^4 (θ) − 9sin^3 (θ) + 14sin^2 (θ) − 9sin(θ) + 2 = 0, for possible values of θ in the range 0 ≤ θ ≤ 2π


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Question Number 12525 by tawa last updated on 24/Apr/17

$Use the substitution t = \sin (θ) to solve the equation$ ${2 \sin}^{4} (θ) - {9 \sin}^{3} (θ) + {14 \sin}^{2} (θ) - 9 \sin (θ) + 2 = 0,$ $for possible values of θ in the range 0 ⩽ θ ⩽ 2 π$
	Answered by mrW1 last updated on 24/Apr/17

	$x = \sin θ$ $2 x^{4} - 9 x^{3} + 14 x^{2} - 9 x + 2 = 0$ $(2 x - 1) (x^{3} - 4 x^{2} + 5 x - 2) = 0$ $(2 x - 1) (x - 2) (x^{2} - 2 x + 1) = 0 x$ $(2 x - 1) (x - 2) {(x - 1)}^{2} = 0$ $\Rightarrow x = \frac{1}{2} \Rightarrow θ = \frac{π}{6}, \frac{5 π}{6}$ $\Rightarrow x = 2 (n o s o l u t i o n, s i n c e - 1 ⩽ x ⩽ 1)$ $\Rightarrow x = 1 \Rightarrow θ = \frac{π}{2}$
	Commented by tawa last updated on 24/Apr/17

	$God bless you sir . Thanks for the help .$
	Answered by sma3l2996 last updated on 24/Apr/17

	$l e t t = s i n θ$ $2 t^{4} - 9 t^{3} + 14 t^{2} - 9 t + 2 = 0$ $(t - 1) (t - 2) ({a t}^{2} + b t + c) = 0$ $\Leftrightarrow (t^{2} - 3 t + 2) ({a t}^{2} + b t + c) = 0$ $\Leftrightarrow {a t}^{4} + {b t}^{3} + {c t}^{2} - 3 {a t}^{3} - 3 {b t}^{2} - 3 c t + 2 {a t}^{2} + 2 b t + 2 c = 0$ $a = 2; 2 c = 2; b - 3 a = - 9$ $a = 2; c = 1; b = - 3$ $s o 2 t^{4} - 9 t^{3} + 14 t^{2} - 9 t + 2 = (t - 1) (t - 2) (2 t^{2} - 3 t + 1) = 0$ $2 t^{2} - 3 t + 1 = 0$ $t = 1 o r t = \frac{1}{2}$ $s i n θ = 1 o r s i n θ = \frac{1}{2} o r s i n θ = 2 a n d t h a t i m p o s s i b l e$ $θ = \frac{π}{2} o r θ = \frac{π}{6} o r θ = \frac{5 π}{6}$
	Commented by tawa last updated on 24/Apr/17

	$God bless you sir . Thanks for the help .$
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