Question

4sin^4 θ + 3sin^(2) θ-1=0

Answer 1

First let x = sin^2(theta)
Now re-write the equation as this:
\[4x^2+3x-1=0\]Now factor this equation:
\[4x^2+4x-x-1=0\]\[4x(x+1)-1(x+1)=0\]\[(4x-1)(x+1)=0\]Now either 4x-1 = 0 or x+1=0. This is the point where you can substitute sin^2(theta) back in for x and solve for the possible x-values.

Case 1 (All values of theta are in radians):
\[4\sin^2(\theta)-1=0\rightarrow \sin(\theta)=\pm\frac{ 1 }{ 2 }\]\[\theta=\pm \frac{ \pi }{ 6 },\pm \frac{ 5\pi }{ 6 }\] Case 2 (All values of theta are in radians):
\[\sin^2(\theta)+1=0\rightarrow \sin(\theta) \notin \mathbb{R} \] Therefore, \[\theta=\pm \frac{ \pi }{ 6 },\pm \frac{ 5\pi }{ 6 }\]

Answer 2

@Sda370

Answer 3

Also, you didn't specify the domain for theta like whether it was [0,2pi] or not so I assumed that the domain isn't restricted.