Answer to Question #156114 in Trigonometry for .

Question

solve the following equation for 0«x«360

a) 6 cos²x + sin x - 4

b) 9 tan x + tan²x + 5 sec² x - 3

Answer 1

a)$6\cos^2 x+\sin x-4=0$

$6(1-\sin^2 x)+\sin x-4=0$

$-6\sin^2 x+\sin x+2=0$

This quadratic equation has solutions $\sin x=-\frac{1}{2}$ and $\sin x=\frac{2}{3}$

The equation $\sin x=-\frac{1}{2}$ has solutions $210^\circ$ and $330^\circ$ in $[0^\circ,360^\circ]$, and since $\frac{2}{3}>0$, the equation $\sin x=\frac{2}{3}$ has solutions $\sin^{-1}\frac{2}{3}$ and $180^\circ-\sin^{-1}\frac{2}{3}$ in $[0^\circ,360^\circ]$

b)$9\tan x+\tan^2 x+5\sec^2 x-3=0$

$9\tan x+\tan^2 x+5(1+\tan^2 x)-3=0$

$6\tan^2 x+9\tan x+2=0$

This quadratic equation has solutions $\tan x=-\frac{9-\sqrt{33}}{2}$ and $\tan x=-\frac{9+\sqrt{33}}{2}$

Since $-\frac{9-\sqrt{33}}{2}<0$, the equation $\tan x=-\frac{9-\sqrt{33}}{2}$ has solutions $360^\circ-\tan^{-1}\frac{9-\sqrt{33}}{2}$ and $180^\circ-\tan^{-1}\frac{9-\sqrt{33}}{2}$ in $[0^\circ,360^\circ]$

Similarly the equation $\tan x=-\frac{9+\sqrt{33}}{2}$ has solutions $360^\circ-\tan^{-1}\frac{9+\sqrt{33}}{2}$ and $180^\circ-\tan^{-1}\frac{9+\sqrt{33}}{2}$

Answer: a)$210^\circ, 330^\circ, \sin^{-1}\frac{2}{3}, 180^\circ-\sin\frac{2}{3}$

b)$360^\circ-\tan^{-1}\frac{9-\sqrt{33}}{2}, 180^\circ-\tan^{-1}\frac{9-\sqrt{33}}{2},$

$360^\circ-\tan^{-1}\frac{9+\sqrt{33}}{2}, 180^\circ-\tan^{-1}\frac{9+\sqrt{33}}{2}$