How do you solve ${sin}^{4} x + 2 {sin}^{2} x - 3 = 0$ $sin^4x+2sin^2x-3=0$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

Question

7387 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-16T21:30:34

$x = \frac{π}{2}, \frac{3 π}{2}$ $x = pi/2, (3pi)/2$

This can be factored as $(x^{2} + a) (x^{2} + b)$ $(x^2 + a)(x^2 + b)$ .

$\Rightarrow ({sin}^{2} x + 3) ({sin}^{2} x - 1) = 0$ $=>(sin^2x + 3)(sin^2x - 1) = 0$

$\Rightarrow {sin}^{2} x = - 3 and {sin}^{2} x = 1$ $=>sin^2x = -3 and sin^2x = 1$

$\Rightarrow sin x = \sqrt{- 3} and sin x = \sqrt{1}$ $=> sinx = sqrt(-3) and sinx = sqrt(1)$

$\Rightarrow sin x = \emptyset and sin x = \pm 1$ $=> sinx = O/ and sinx = +-1$

$\Rightarrow x = \frac{π}{2}, \frac{3 π}{2}$ $=> x = pi/2, (3pi)/2$

Hopefully this helps!

How do you solve sin^4x+2sin^2x-3=0sin4x+2sin2x−3=0sin^4x+2sin^2x-3=0 for 0<=x<=2pi0≤x≤2π0<=x<=2pi?