Umm, these are really very hard to solve. Need tricks.
Even when I was yet in High School, I resort to iteration, or trial and error, in solving equations that I didn't know how to solve the usual ways.
Problem 1 is is a good candidate for my iteration.
By luck, or because of the sqrt(3) in the equation, I tried first, x = 30degrees.
cos(3*30deg) =? 2cos(30deg) -sqrt(3)*sin(3*30deg)
cos(90deg) =? 2cos(30deg) -sqrt(3)*sin(90deg)
0 =? 2[sqrt(3) / 2] -sqrt(3)*(1)
0 =? sqrt(3) -sqrt(3)
0 =? 0
Yes, so x = 30deg is OK.
I was just lucky to hit it in one try only. So, 30deg is a solution. There might be some more but I don't know how, or I don't want, to continue after that.
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For the second problem, another trick.
cos(2x +60deg) = sin(x) ----------(ii)
Expanding that will lead to an equation that is as complicated as the first one, so another dip into our bag of tricks:
We know that cosA = sin(90deg -A).
That means A +(90deg -A) = 90 deg.
Or, cosA = sinB -----if angles A and B are complementary.
So, using that on (ii),
(2x +60deg) +x = 90deg
3x = 90 -60
x = 30/3
x = 10 deg ------------a solution.
Check,
cos(2*10deg +60deg) =? sin(10deg)
cos(80deg) =? sin(10deg)
Using a calculator,
0.173648178 =? 0.173648178
Yes, so, x = 10 degrees is correct.