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

4cos ^2 x sin x −2sin ^2 x = 3sin x  where −(π/2)≤x≤(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)) ⊸◊⊸)) :}

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))}

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))}