Given that cos^2 x = 6 sin^4 x, find sin 2x.

Step-by-step solution

1. 1

   Reduce everything to one quantity. Both sides involve only even powers, so the Pythagorean identity lets us work with a single unknown. Put

   $u=\sin^{2}x,\qquad\text{so}\qquad\cos^{2}x=1-u,\qquad \sin^{4}x=u^{2}.$

   Choosing $u=\sin^{2}x$ rather than $\sin x$ avoids sign complications entirely — nothing in the problem distinguishes $x$ from $-x$.

2. 2

   Form and solve the quadratic. Substituting into $\cos^{2}x=6\sin^{4}x$:

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

   so $u=\tfrac13$ or $u=-\tfrac12$.

3. 3

   Discard the impossible root. Since $u=\sin^{2}x$ is a square, $u\ge0$, so $u=-\tfrac12$ is rejected. Moreover $u=\sin^{2}x\le1$, and $\tfrac13$ passes that test too. Hence

   $\sin^{2}x=\frac13,\qquad \cos^{2}x=1-\frac13=\frac23.$

   Checking the original condition: $6\sin^{4}x=6\cdot\tfrac19=\tfrac23=\cos^{2}x$ ✓.

4. 4

   Use the double-angle formula in squared form. From $\sin 2x=2\sin x\cos x$, squaring avoids ever needing the individual signs of $\sin x$ and $\cos x$:

   $\sin^{2}2x=4\sin^{2}x\cos^{2}x=4\cdot\frac13\cdot\frac23=\frac89.$

5. 5

   Take the square root and keep both signs. Therefore

   $\sin 2x=\pm\sqrt{\frac89}=\pm\frac{2\sqrt2}{3}\approx\pm0.9428.$

   Both signs genuinely occur: the original condition constrains only $\sin^{2}x$, so $x$ may lie in any quadrant, and $\sin 2x$ is positive in the first and third quadrants and negative in the second and fourth. Without extra information about the quadrant, the answer must carry the $\pm$.