1. ## Trigonometric equation

if $\sin x+\tan x=\frac{5}{6}$ then $\sin x\tan x=$

2. ## Re: Trigonometric equation

What steps have you taken to try and solve this ?

3. ## Re: Trigonometric equation

Originally Posted by jacks
if $\sin x+\tan x=\frac{5}{6}$ then $\sin x\tan x=$
Dear jacks,

$\sin x+\tan x=\frac{5}{6}$

$\sin x\left(1+\frac{1}{\cos x}\right)=\frac{5}{6}$

$\pm\sqrt{1-\cos^{2}x}\left(\frac{1+\cos x}{\cos x}\right)=\frac{5}{6}$

$(1-\cos^{2}x)\left(\frac{1+\cos x}{\cos x}\right)^2=\frac{25}{36}$

$1-2\cos x-\frac{25}{36}\cos^{2}x+2\cos^{3}x-\cos^{4}x=0$

By solving this quartic equation you could obtain,

$\cos x=\begin{cases}~~0.91710326\\-0.50822898\\-1.2044371+0.83354720i\\-1.2044371-0.83354720i\end{cases}$

$\sin^{2}x+\tan^{2}x=(1-\cos^{2}x)+\left(\frac{1}{\cos^{2}x}-1\right)=\left\{\begin{array}{l}~~0.347871430\\~~3 .61322011\\-0.59165723+2.44412431i\\-0.59165723-2.44412431i\end{array}\right.$

Also note that,

$(\sin x+\tan x)^2=\frac{25}{36}$

$\sin^{2}x+\tan^{2}x+2\sin x\tan x=\frac{25}{36}$

$\sin x\tan x=\frac{25}{72}-\frac{\left(\sin^{2}x+\tan^{2}x\right)}{2}$

$\sin x\tan x=\begin{cases}~~0.17328651\\-1.45938783\\~~0.64305084-1.22206216i\\~~0.64305084+1.22206216i\end{cases}$