Question

Trigonometry-Logarithm Question..Please help..

Answer 1

`$x=(25\pi )/(12) \text{ and }x=(29\pi )/(12)`

Explanation:

Solve

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

in interval
`$\left(2\pi, (5\pi)/(2) \right)$`

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`

`$(1)/(2) =\sin 2x$`

The solutions for it are:

`$2x=(\pi )/(6)+2\pi n \text{ and }2x=(5\pi )/(6)+2\pi n, \text{ as } n\in \mathbb{Z}$`

As

`$x=(\pi )/(12)+\pi n \text{ and }x=(5\pi )/(12)+\pi n, \text{ as } n\in \mathbb{Z}$`

But we have the interval

`$\left(2\pi, (5\pi)/(2) \right)$`

Therefore, the answer is

`$x=(25\pi )/(12) \text{ and }x=(29\pi )/(12)`