We start with polar coordinates of a point $P=(r,\theta)$ vs its Cartesian form $(x,y)$. You know $x=r\cos\theta, y=r\sin\theta$. Now rotate the point $P$ by an angle of $\alpha$ with respect to origin so its angle with respect to $x$ axis becomes $\theta+\alpha$. Its new polar coordinates is $P'=(r,\theta+\alpha)$ and its new Cartesian coordinates $(x',y')$, where $x'=r\cos(\theta+\alpha), y'=r\sin(\theta+\alpha)$. Use trig identities to expand these
$x'=r(\cos \theta \cos \alpha -\sin \theta \sin \alpha)=x\cos \alpha -y \sin\alpha$, and
$ y'=r(\sin\theta\cos\alpha+\cos\theta\sin\alpha)=y\cos\alpha+x\sin\alpha$.
If you write these as a matrix equation you can finish your problem.
$
\left[ \begin{array}{c} x' \\ y' \end{array} \right] = \begin{bmatrix} \cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha \\ \end{bmatrix} \times \left[ \begin{array}{c} x \\ y \end{array} \right]
$
If you have several rotation $\alpha_1, \cdots, \alpha_n$ then the effect of multiplying all the related matrices is the same as adding the given angles. In particular if you have $n$ rotations by the same angle $\alpha$ the result will be same as one rotation of $n\alpha$. In short
$
\begin{bmatrix} \cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha \\ \end{bmatrix}^n= \begin{bmatrix} \cos n\alpha & -\sin n\alpha \\
\sin n\alpha & \cos n\alpha \\ \end{bmatrix}$
You have
$$(18\log_32\log_43-3\log_32\log_43)(2\log_23\log_32-2\log_32\log_83)\;,$$
which is correct. Clearly
$$18\log_32\log_43-3\log_32\log_43=15\log_32\log_43\;,$$
so we can immediately simplify it to
$$15\log_32\log_43(2\log_23\log_32-2\log_32\log_83)\;.\tag{1}$$
Now if $4^x=3$, then $2^{2x}=3$, so $\log_23=2\log_43$, or $\log_43=\frac12\log_23$. Similarly, you can verify that $\log_83=\frac13\log_23$. Thus, we can further simplify $(1)$ to
$$\frac{15}2\log_32\log_23\left(2\log_23\log_32-\frac23\log_32\log_23\right)\;.\tag{2}$$
This is extremely easy to evaluate if you know something about products of the form $\log_ab\log_ba$. If not, use the fact that in general $\log_bx=\frac{\log_ax}{\log_ab}$. I’ll complete the calculation below but leave it spoiler-protected; mouse-over to see it.
A useful general fact is that $\log_ab\log_ba=1$; this follows easily from the fact that in general $\log_bx=\frac{\log_ax}{\log_ab}$. Thus, $\log_23\log_32=1$, and $(2)$ is simply $\dfrac{15}2\left(2-\dfrac23\right)=\dfrac{15}2\cdot\dfrac43=10$.
Best Answer
We need to solve $\cos(x) + 2\sin(x) = 0$ and $\cos(x) - \sin(x) = 0$. The first equation is solved by setting $u = \cos(x)$ and solving for $u$, i.e. \begin{align*} u+2\sqrt{1-u^2} &= 0\\ u &= -2\sqrt{1-u^2} \\ u^2 &= 4(1-u^2) \\ 5u^2 &= 4 \end{align*} and checking which values of $\cos^{-1}(\pm 2/\sqrt{5})$ lie in the specified interval. For the second equation, only $x = \pi/4$ works.