Solve 6 sin to the fourth power of x, plus sin squared x, minus 6 equals 0.

Step-by-step solution

1. 1

   Substitute to expose a quadratic. Only even powers of $\sin x$ appear, so set $u=\sin^2x$:

   $6u^2+u-6=0$

   The substitution also records a constraint that the quadratic itself does not know about: $u=\sin^2x$ forces $0\le u\le 1$.

2. 2

   Solve the quadratic. With $\Delta=1^2-4(6)(-6)=1+144=145$ (not a perfect square),

   $u=\frac{-1\pm\sqrt{145}}{12}$

3. 3

   Discard the inadmissible root. Since $\sqrt{145}\approx 12.0416$,

   $u_1=\frac{-1+\sqrt{145}}{12}\approx 0.9201329,\qquad u_2=\frac{-1-\sqrt{145}}{12}\approx -1.0868$

   Only $u_1$ satisfies $0\le u\le 1$; $u_2$ is negative and a square can never be negative, so it is rejected. (Had $u_1$ exceeded $1$ the equation would have had no solution at all.)

4. 4

   Take square roots to get $\sin x$.

   $\sin x=\pm\sqrt{\frac{\sqrt{145}-1}{12}}\approx \pm 0.9592356$

   Both signs must be kept, because $\sin^2 x$ loses the sign information.

5. 5

   Write the general solution. The two sign choices combine into the standard family for a sine equation:

   $x=n\pi\pm\arcsin\!\left(\sqrt{\frac{\sqrt{145}-1}{12}}\right),\qquad n\in\mathbb{Z}$

   Numerically $\arcsin(0.9592356)\approx 1.2843$ rad $\approx 73.58^{\circ}$, so the solutions are $x\approx n\pi\pm 1.2843$.

6. 6

   Verify. With $u=0.9201329$: $6u^2+u-6=6(0.8466445)+0.9201329-6=5.0798671+0.9201329-6=0$ ✓. And taking $x=1.2843$, $\sin x=0.9592356$, so $\sin^2x=u$ as required.