Given 6 sin^4 x + sin^2 x - 6 = 0, find the value of 2 sin x cos x.

Step-by-step solution

1. 1

   Turn the trigonometric equation into a quadratic. Only even powers of $\sin x$ appear, so set $u=\sin^{2}x$; then $\sin^{4}x=u^{2}$ and

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

   Note the target $2\sin x\cos x=\sin 2x$ also squares to something expressible in $u$, so no individual value of $x$ will be needed.

2. 2

   Solve the quadratic and discard the impossible root.

   $u=\frac{-1\pm\sqrt{1+144}}{12}=\frac{-1\pm\sqrt{145}}{12}$

   Since $u=\sin^{2}x$ must satisfy $0\le u\le 1$, the negative root $\frac{-1-\sqrt{145}}{12}\approx-1.09$ is rejected. That leaves

   $\sin^{2}x=\frac{\sqrt{145}-1}{12}\approx 0.92013$

   which is safely inside $[0,1]$ ✓.

3. 3

   Get $\cos^{2}x$ from the Pythagorean identity.

   $\cos^{2}x=1-\frac{\sqrt{145}-1}{12}=\frac{13-\sqrt{145}}{12}\approx 0.07987$

4. 4

   Square the target instead of computing it directly. Squaring avoids ever needing the individual signs of $\sin x$ and $\cos x$:

   $\left(2\sin x\cos x\right)^{2}=4\sin^{2}x\cos^{2}x=4\cdot\frac{\sqrt{145}-1}{12}\cdot\frac{13-\sqrt{145}}{12}=\frac{\left(\sqrt{145}-1\right)\left(13-\sqrt{145}\right)}{36}$

5. 5

   Expand the surd product.

   $\left(\sqrt{145}-1\right)\left(13-\sqrt{145}\right)=13\sqrt{145}-145-13+\sqrt{145}=14\sqrt{145}-158$

   so

   $\left(2\sin x\cos x\right)^{2}=\frac{14\sqrt{145}-158}{36}=\frac{7\sqrt{145}-79}{18}\approx 0.29395$

6. 6

   Take the square root and keep both signs.

   $2\sin x\cos x=\sin 2x=\pm\sqrt{\frac{7\sqrt{145}-79}{18}}\approx\pm 0.5422$

   Both signs occur: the original equation constrains only $\sin^{2}x$, so $x$ may lie in a quadrant where $\sin x$ and $\cos x$ have the same sign or opposite signs. Numerically $\sin^{2}x=0.920133$ gives $6(0.846645)+0.920133-6=0.000000$ ✓.