# How do you find the solution for 6 tan^2 x - 4 sin^2 x = 1 for [0,360]?

Oct 1, 2015

S = {30º,150º,210º,330º}

#### Explanation:

Multiply both sides by ${\cos}^{2} x$

$6 {\sin}^{2} x - 4 {\sin}^{2} x {\cos}^{2} x = {\cos}^{2} x$

Use the pythagorean identity ${\sin}^{2} x = 1 - {\cos}^{2} x$

$6 \left(1 - {\cos}^{2} x\right) - 4 \left(1 - {\cos}^{2} x\right) {\cos}^{2} x = {\cos}^{2} x$

Simplify

$6 - 6 {\cos}^{2} x - 4 {\cos}^{2} x + 4 {\cos}^{4} x = {\cos}^{2} x$
$6 - 10 {\cos}^{2} x + 4 {\cos}^{4} x = {\cos}^{2} x$

Subsitute ${\cos}^{2} x$ for $z$

$6 - 10 z + 4 {z}^{2} = z$

Make one side equal zero

$4 {z}^{2} - 11 z + 6 = 0$

$z = \frac{11 \pm \sqrt{121 - 4 \cdot 4 \cdot 6}}{8}$
$z = \frac{11 \pm \sqrt{25}}{8} = \frac{11 \pm 5}{8}$
${z}_{+} = \frac{11 + 5}{8} = \frac{16}{8} = 2$
${z}_{-} = \frac{11 - 5}{8} = \frac{6}{8} = \frac{3}{4}$

Subsitute $z$ for ${\cos}^{2} x$

${\cos}^{2} \left(x\right) = 2$ is invalid because the square cosine has a range of $0 \le y \le 1$

Which leaves us with ${\cos}^{2} x = \frac{3}{4} \rightarrow \cos x = \pm \frac{\sqrt{3}}{2}$

The cosine has one of those values on the first quadrant when theta = 30º, one of those values on the second quadrant when theta = 180º - 30º = 150º, one of those values on the third quadrant when theta = 180º+30º = 210º and one of the values on the fourth quadrant when theta = 360º - 30º = 330º

So, the set of solutions $S$ is S = {30º,150º,210º,330º}