Given that cos x = sqrt(6) * sin^2 x, find sin 2x.

Step-by-step solution

1. 1

   Note the sign information hidden in the hypothesis. The right side $\sqrt6\sin^{2}x$ is a non-negative multiple of a square, so immediately

   $\cos x\ge0.$

   This observation, made before any algebra, is what will let us pin the sign of $\cos x$ later — the squared version of this problem loses it.

2. 2

   Square both sides to get a single-variable equation. Squaring gives $\cos^{2}x=6\sin^{4}x$, and substituting $u=\sin^{2}x$ with $\cos^{2}x=1-u$:

   $1-u=6u^{2}\;\Longrightarrow\;6u^{2}+u-1=0\;\Longrightarrow\;(3u-1)(2u+1)=0.$

   Squaring can introduce extraneous solutions, so the sign check from step 1 will have to be applied at the end.

3. 3

   Select the admissible root. The roots are $u=\tfrac13$ and $u=-\tfrac12$; the latter is impossible since $u=\sin^{2}x\ge0$. Hence

   $\sin^{2}x=\frac13,\qquad \sin x=\pm\frac{\sqrt3}{3}.$

4. 4

   Recover $\cos x$ from the original equation, not from the squared one. Substituting $\sin^{2}x=\tfrac13$ back into the given relation:

   $\cos x=\sqrt6\cdot\frac13=\frac{\sqrt6}{3}\approx0.8165,$

   which is positive, consistent with step 1 — so no extraneous root has crept in. Check: $\sin^{2}x+\cos^{2}x=\tfrac13+\tfrac69=\tfrac13+\tfrac23=1$ ✓.

5. 5

   Apply the double-angle formula. With $\sin x=\pm\tfrac{\sqrt3}{3}$ and $\cos x=+\tfrac{\sqrt6}{3}$:

   $\sin 2x=2\sin x\cos x=2\left(\pm\frac{\sqrt3}{3}\right)\frac{\sqrt6}{3}=\pm\frac{2\sqrt{18}}{9}=\pm\frac{6\sqrt2}{9}=\pm\frac{2\sqrt2}{3}\approx\pm0.9428.$

   Both signs remain because $\cos x>0$ only rules out two of the four quadrants; $x$ may still lie in the first or the fourth, giving $\sin 2x$ positive or negative respectively.