Question

Solve the equation: sin^-1 (4x^4+x^2) = 1/6 pi

Answer 1

sin^-1 (4x^4+x^2) = 1/6 pi
4x^4+x^2 = sin(pi/6) = 1/2
8x^4+2x^2-1 = 0
x^2 = [-2±√(4+32)]/16
= [-2±6]/16
= -1/2 or 1/4

x^2 = -1/2 x = ±i/√2
x^2 = 1/4 means x = ±1/2

Answer 2

Sin(pi/6) does not = 1/2

Answer 3

Sin^-1(4x⁴+x²)=1/6 pi
(4x⁴+x²) = Sin pi/6
4x⁴+x² = 0.5 (put your calculator in radian mode)
Then, 4x⁴+x²-0.5=0
Multiply this equ by 2 we get,
8x⁴+2x²-1=0
Disguised equ, let t=x²
We get, 8t²+2t-1=0
Sum: 4, Product:-8
(4-2) (4×-2)
Then, 8t²+2t-1=0
8t²+4t-2t+1=0
4t(2t+1)-1(2t+1)=0
(4t-1)(2t+1) =0
Replace t by x²
We get 4x²-1=0 or 2x²+1=0
4x² =1 2x² =-1
x² = 1/4 x² =-1/2
x =±1/2 x = ± i/√2