Question

The solution of equation $5\sin^{2}x - 7\sin x\cos x + 16\cos^{2}x = 4$ is

• A. $x = n\pi + \tan^{- 1}3$or$x = n\pi + \tan^{- 1}4$
• B. $x = n\pi + \frac{\pi}{6}$or$x = n\pi + \frac{\pi}{4}$
• C. $x = n\pi$or$x = n\pi + \frac{\pi}{4}$
• D. None of these

Answer 1

Answer: (A)

$x = n\pi + \tan^{- 1}3$or$x = n\pi + \tan^{- 1}4$

Answer 2

To solve this kind of equation; we use the fundamental formula trigonometrical identity, $\sin^{2}x + \cos^{2}x = 1$

writing the equation in the form,

$5\sin^{2}x - 7\sin x\cos x + 16\cos^{2}x = 4(\sin^{2}x + \cos^{2}x)$

$\Rightarrow$ $\sin^{2}x - 7\sin x\cos x + 12\cos^{2}x = 0$Dividing by $\cos^{2}x$ on

both sides we get, $\tan^{2}x - 7\tan x + 12 = 0$

Now it can be factorized as; $(\tan x - 3)(\tan x - 4) = 0$

$\Rightarrow$ $\tan x = 3,4$

i.e., $\tan x = \tan(\tan^{- 1}3)$ or $\tan x = \tan(\tan^{- 1}4)$

$\Rightarrow$ $x = n\pi + \tan^{- 1}3$ or $x = n\pi + \tan^{- 1}4.$