Question

Trigonometry-Logarithm Question..Please help..

Answer 1

Answer:

$x=\frac{25\pi }{12} \text{ and }x=\frac{29\pi }{12}$

Step-by-step explanation:

Solve

$\log_2(2 \sin x) + \log_2(\cos x) =1$

in interval $\left(2\pi, \frac{5\pi}{2} \right)$

Remember that logarithms follow:

$f(x\cdot y)=f(x)+ f(y)$

Therefore

$\log_2(2 \sin x) + \log_2(\cos x) =-1 \implies \log_2(2 \sin x \cdot \cos x) =-1$

We got a double angle of sine:

$\sin(2x)= 2 \sin x \cos x$

$\log_2( \sin 2x ) =-1$

$2^{-1}=\sin 2x$

$\frac{1}{2} =\sin 2x$

The solutions for it are:

$2x=\frac{\pi }{6}+2\pi n \text{ and }2x=\frac{5\pi }{6}+2\pi n, \text{ as } n\in \mathbb{Z}$

As

$x=\frac{\pi }{12}+\pi n \text{ and }x=\frac{5\pi }{12}+\pi n, \text{ as } n\in \mathbb{Z}$

But we have the interval

$\left(2\pi, \frac{5\pi}{2} \right)$

Therefore, the answer is

$x=\frac{25\pi }{12} \text{ and }x=\frac{29\pi }{12}$