4cos^2 x sin x −2sin^2 x = 3sin x where −(π/2)≤x≤(π/2)


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Question Number 104904 by bemath last updated on 24/Jul/20

$4 \cos^{2} x \sin x - 2 \sin^{2} x = 3 \sin x$ $w h e r e - \frac{π}{2} ⩽ x ⩽ \frac{π}{2}$
	Answered by bramlex last updated on 24/Jul/20
	$sin x (4cos^2 x−2sin x−3) = 0 { ((sin x = 0 → x_1 = 0)),((4cos^2 x −2sin x−3 = 0)) :} 4−4sin^2 x−2sin x−3 = 0 4sin^2 x+2sin x−1 = 0 sin x = ((−2 ± (√(4+16)))/8) = ((−1±(√5))/4) { ((x_2 = 18° = (π/(10)))),((x_3 = −54°= −((3π)/(10)) ⊸◊⊸)) :}$
	$\sin x (4 \cos^{2} x - 2 \sin x - 3) = 0$ ${\begin{cases} \sin x = 0 \to x_{1} = 0 \\ {4 \cos}^{2} x - 2 \sin x - 3 = 0 \end{cases}$ $4 - 4 \sin^{2} x - 2 \sin x - 3 = 0$ $4 \sin^{2} x + 2 \sin x - 1 = 0$ $\sin x = \frac{- 2 \pm \sqrt{4 + 16}}{8} = \frac{- 1 \pm \sqrt{5}}{4}$ ${\begin{cases} x_{2} = 18 ° = \frac{π}{10} \\ x_{3} = - 54 ° = - \frac{3 π}{10} ⊸ ◊ ⊸ \end{cases}$
	Answered by Dwaipayan Shikari last updated on 24/Jul/20
	$2sinx(2cos^2 x−sinx)=3sinx 2−2sin^2 x−sinx=(3/2) or sinx=0 2sin^2 x+sinx−(1/2)=0 4sin^2 x+2sinx−1=0 sinx=((−2+(√(4+16)))/8)=((−(√5)±1)/4) or sinx=0 ⇒x=kπ sinx=sin(π/(10)) ( k∈Z) x=kπ±(π/(10)) or sinx=−((((√5)+1)/4)) x=−((3π)/(10)) Solutions are {(π/(10)),0,−((3π)/(10))}$
	$2 s i n x (2 {c o s}^{2} x - s i n x) = 3 s i n x$ $2 - 2 {s i n}^{2} x - s i n x = \frac{3}{2} o r s i n x = 0$ $2 {s i n}^{2} x + s i n x - \frac{1}{2} = 0$ $4 {s i n}^{2} x + 2 s i n x - 1 = 0$ $s i n x = \frac{- 2 + \sqrt{4 + 16}}{8} = \frac{- \sqrt{5} \pm 1}{4} o r s i n x = 0 \Rightarrow x = k π$ $s i n x = s i n \frac{π}{10} (k \in Z)$ $x = k π \pm \frac{π}{10}$ $o r s i n x = - (\frac{\sqrt{5} + 1}{4}) x = - \frac{3 π}{10}$ $S o l u t i o n s a r e {\frac{π}{10}, 0, - \frac{3 π}{10}}$
	Answered by 1549442205PVT last updated on 24/Jul/20
	$⇔sinx(4cos^(2 ) x−2sinx−3)=0 ⇔sinx[(4(1−sin^(2 ) x)−2sinx−3]=0 ⇔sinx(4sin^2 x+2sinx−1)=0 i)sinx=0⇔x=kπ .Since x∈[−(π/2),(π/2)] ⇒x=0 ii)4sin^2 x+2sinx−1=0 ⇔sinx =((−1+(√5))/4) or sinx=((−1−(√5))/4) a)sinx=(((√5)−1)/4)=sin(π/(10))⇒x=(π/(10)) b)sinx=((−((√5)+1))/4)=sin((−3π)/(10))⇒x=((−3π)/(10)) Thus,x∈{0,(π/(10)),((−3π)/(10))}$
	$\Leftrightarrow sinx ({4 \cos}^{2} x - 2 sinx - 3) = 0$ $\Leftrightarrow sinx [(4 (1 - \sin^{2} x) - 2 sinx - 3] = 0$ $\Leftrightarrow sinx ({4 \sin}^{2} x + 2 sinx - 1) = 0$ $i) sinx = 0 \Leftrightarrow x = k π . Since x \in [- \frac{π}{2}, \frac{π}{2}]$ $\Rightarrow x = 0$ $ii) {4 \sin}^{2} x + 2 sinx - 1 = 0$ $\Leftrightarrow sinx = \frac{- 1 + \sqrt{5}}{4} or sinx = \frac{- 1 - \sqrt{5}}{4}$ $a) sinx = \frac{\sqrt{5} - 1}{4} = \sin \frac{π}{10} \Rightarrow x = \frac{π}{10}$ $b) sinx = \frac{- (\sqrt{5} + 1)}{4} = \sin \frac{- 3 π}{10} \Rightarrow x = \frac{- 3 π}{10}$ $Thus, x \in {0, \frac{π}{10}, \frac{- 3 π}{10}}$
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