first question i finished, but im not sure if notation is correct? second one i couldn't finish. third one i don't know if i can even start because of typo on paper
1.
$$ \cos^2(x)-0.2\sin(x) = 0.9; \\ [0^{\circ},360^{\circ})$$
use identity
$(1-\sin^2(x))-\frac{1}{5}\sin(x) = \frac{9}{10}$ $-\sin^2(x)-\frac{1}{5}\sin(x) = \frac{9}{10} - \frac{10}{10}$ $-\sin^2(x)-\frac{1}{5}\sin(x) + \frac{1}{10} = 0$
negate to factor
$$\sin^2(x)+\frac{1}{5}\sin(x) - \frac{1}{10} = 0$$
can't factor, use quadratic formula
$$A=1,B=\frac{2}{10},C=-\frac{1}{10}$$ $$\sin(x) = \frac{-\frac{2}{10}\pm \sqrt{\frac{4}{100}+(\frac{4}{10})}}{2}$$
$$\sin(x) = \frac{-\frac{2}{10}\pm \sqrt{\frac{11}{25}}}{2}$$
$$\sin(x) = 0.231662479$$ $$\sin(x) = -0.431662479$$
$$\sin^{-1}(0.231662479) = 13.39496861 $$ $$\sin^{-1}(-0.431662479) = -25.57311158$$ $$x = 13.39496861^{\circ}$$ $$x = -25.57311158^{\circ}$$
final answer
$$x = 13.39496861^{\circ} + k180^{\circ}, k \text{ any integer}$$ $$x = -25.57311158^{\circ} + k180^{\circ}, k \text{ any integer}$$
correct? this is how my book writes it
2.
$$\sin(2x)\cos(x)+\cos(2x)\sin(x)=\cos(3x) \quad\forall x \in \mathbb{R}$$
use identity $\sin(3x)$
$$\sin(3x)=\cos(3x)$$
$$\tan(3x)=1$$
$$\tan^{-1}(1)=.785398163 \rm{rad}$$ $$ x = \frac{.785398163}{3}$$
final answer
$$x=.261799388rad +\pi k, k \text{ any integer}$$
3.
this is exactly how the problem is written on the paper, this must be a typo since there is no equal sign? I think the plus sign was meant to be a equal sign
Hint
For 2 $$\sin(2x)\cos(x)+\cos(2x)\sin(x)=\sin(2x+x)=\sin(3x)$$ $$\cos(2x)\cos(x)-\sin(2x)\sin(x)=\cos(2x+x)=\cos(3x)$$
I am sure that you can take from here.