Question

Solve this 12th question ASAP​

Answer 1

Step-by-step explanation:

We have,

$4 \sin(x) \cos(x) + 2 \sin(x) + 2 \cos(x) + 1 = 0$

$\implies \: 2 \sin(x) \{2 \cos(x) + 1 \} + 1 \{2 \cos(x) + 1 \} = 0$

$\implies \: \{ 2 \sin(x) + 1 \} \{2 \cos(x) + 1 \} = 0$

$\implies \: 2 \sin(x) + 1 = 0 \: \: \: \: or \: \: \: \: 2 \cos(x) + 1 = 0$

$\implies \: \sin(x) = - \frac{1}{2} \: \: \: \: or \: \: \: \: \cos(x) = - \frac{1}{2}$

$\implies \: \sin(x) = \sin \bigg(-\frac{\pi}{6} \bigg) \: \: \: \: or \: \: \: \: \cos(x) = \cos \bigg(\frac{2\pi}{3} \bigg)$

$\implies \:x= n\pi-( - 1)^{n} .\frac{\pi}{6} \: \: \: \: or \: \: \: \: x = 2m\pi \pm\frac{2\pi}{3}$

So,

$x = \frac{7\pi}{6} , \frac{11\pi}{6} \: \: \: \: \: \: or \: \: \: \: \: \: x = \frac{2\pi}{3}, \frac{4\pi}{3}$