Question

If $3\sin^2x-8\sin x+4=0$, $x\in\left(\dfrac{\pi}{2},\pi\right)$, then $\tan x=$

• A. $-\dfrac{\sqrt5}{2}$
• B. $\dfrac{2}{\sqrt5}$
• C. $-\dfrac{2}{\sqrt5}$
• D. $\dfrac{\sqrt5}{2}$

Hint:

Always use the given interval to determine the correct sign of trigonometric ratios.

Answer 1

The Correct Option is C

Step 1: Solving the given quadratic in $\sin x$.
\[ 3\sin^2x-8\sin x+4=0 \] \[ (3\sin x-2)(\sin x-2)=0 \] \[ \sin x=\frac{2}{3} \] 
Step 2: Identifying the quadrant. 
Since $x\in\left(\dfrac{\pi}{2},\pi\right)$, $\sin x>0$ and $\cos x<0$. 
Step 3: Finding $\tan x$. 
\[ \cos x=-\sqrt{1-\sin^2x} =-\sqrt{1-\frac{4}{9}}=-\frac{\sqrt5}{3} \] \[ \tan x=\frac{\sin x}{\cos x} =\frac{\frac{2}{3}}{-\frac{\sqrt5}{3}} =-\frac{2}{\sqrt5} \] 
Step 4: Conclusion. 
The value of $\tan x$ is $-\dfrac{2}{\sqrt5}$.