Question

A question includes logarithm and trigonometry. Could anybody help me to solve this,please?

Answer 1

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

Condense the logarithms on the left side:

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

Recall the double angle identity for sine,
`\sin(2x)=2\sin x\cos x`:

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

Write both sides as powers of 2:

`2^(\log_2(\sin(2x)))=2^(-1)`

Simplify this:

`\sin(2x)=\frac12`

Solve for
`2x`:

`2x=\sin^(-1)\left(\frac12\right)+2n\pi\text{ OR }2x=\pi-\sin^(-1)\left(\frac12\right)+2n\pi`

(where
`n` is any integer)

Recall that
`\sin^(-1)\left(\frac12\right)=\frac\pi6`:

`2x=\frac\pi6+2n\pi\text{ OR }2x=\frac{5\pi}6+2n\pi`

Solve for
`x`:

`x=\frac\pi{12}+n\pi\text{ OR }x=(5\pi)/(12)+n\pi`

We get solutions in the interval
`2\pi <x<\frac{5\pi}2` when
`n=2`, giving

`\boxed{x=(25\pi)/(12)}\text{ OR }\boxed{x=(29\pi)/(12)}`