Question

Picture below

Answer 1

Please reattach with different names

Answer 2

I do not know what quadrant x lies in, no.

Answer 3

First time ever seeing this.

Answer 4

\[sinx-\sqrt{3-3\sin^2x}=0 \\ \\ sinx-\sqrt{3(1-\sin^2x)}=0\]
Using \(1-sin^2x=cos^2x\)
\[sinx-\sqrt{3(\cos^2x)}=0\]
\[sinx-\sqrt3cosx=0 \\ \\ tanx=\sqrt3\]

Answer 5

Then find the inverse of tangent to get x

Answer 6

Ugh that's confusing. All i know from this stuff is the 30 60 90 triangle

Answer 7

\[tanx=\sqrt3=\frac{\sqrt3}{1}=\frac{opposite}{adjacent}\]