Question

Save me the headache

Answer 1

Simple tell everyone around you to be quiet and to keep quiet

if u have a headache

Explanation:

Answer 2

`(9\sin2x+9\cos2x)^2=81`

Taking the square root of both sides gives two possible cases,

`9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1`

or

`9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1`

Recall that

`\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta`

If
`\alpha=2x` and
`\beta=\frac\pi4`, we have

`\sin\left(2x+\frac\pi4\right)=(\sin2x+\cos2x)/(\sqrt2)`

so in the equations above, we can write

`\sin2x+\cos2x=\sqrt2\sin\left(2x+\frac\pi4\right)=\pm1`

Then in the first case,

`\sqrt2\sin\left(2x+\frac\pi4\right)=1\implies\sin\left(2x+\frac\pi4\right)=\frac1{\sqrt2}`

`\implies2x+\frac\pi4=\frac\pi4+2n\pi\text{ or }\frac{3\pi}4+2n\pi`

(where
`n` is any integer)

`\implies2x=2n\pi\text{ or }\frac\pi2+2n\pi`

`\implies x=n\pi\text{ or }\frac\pi4+n\pi`

and in the second,

`\sqrt2\sin\left(2x+\frac\pi4\right)=-1\implies\sin\left(2x+\frac\pi4\right)=-\frac1{\sqrt2}`

`\implies2x+\frac\pi4=-\frac\pi4+2n\pi\text{ or }-\frac{3\pi}4+2n\pi`

`\implies2x=-\frac\pi2+2n\pi\text{ or }-\pi+2n\pi`

`\implies x=-\frac\pi4+n\pi\text{ or }-\frac\pi2+n\pi`

Then the solutions that fall in the interval
`[0,2\pi)` are

`x=0,\frac\pi4,\frac\pi2,\frac{3\pi}4,\pi,\frac{5\pi}4,\frac{3\pi}2,\frac{7\pi}4`