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The solution by tutor @Theo is incomplete (and, therefore, is incorrect).
The method which he uses has a hole, which makes it dangerous.
Therefore, I came to provide a correct solution.
Solution
5*sin(2x) = -7.5*cos(x). (1)
Use a general trigonometry identity for sine of the double angle
sin(2x) = 2*sin(x)*cos(x).
Substitute it into the original equation (1) to get
5*2*sin(x)*cos(x) = -7.5*cos(x)
10*sin(x)*cos(x) + 7.5*cos(x) = 0 (2)
Factor left side
10*cos(x)*(sin(x) + 0.75) = 0. (3)
Divide both sides by 10
cos(x)*(sin(x) + 0.75) = 0. (4)
Notice that at every step my transformations were equivalent, so equation (1) is EQUIVALENT to equation (4).
In other words, equations (1) and (4) have the same set of solutions.
Now, equation (4) deploys in two independent equations:
(a) cos(x) = 0, and
(b) sin(x) + 0.75 = 0, or, equivalently, sin(x) = -0.75.
It means that the solution set to equation (4) is the union of solutions to equations (a) and (b).
For equation (a) the solutions in the given interval are x= arccos(0) = = 1.5708 and x= arccos(0) + = = 4.7124.
For equation (b) the solutions in the given interval are x= arcsin(-0.75) + = 5.4351 and x = arcsin(0.75) + pi = 3.9897.
ANSWER. The given equation has 4 solutions in the interval [,). They are
x= = 1.5708, x= = 4.7124, x= arcsin(-0.75) + = 5.4351 and x= arcsin(0.75) + = 3.9897.
Numerical values are in radians and are correct with 4 decimal places, as the problem does request.
The solution is completed.
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The lessons to learn from my solution are THESE :
(a) You can not divide equations (2) and (3) by cos(x) without analyzing the case cos(x) = 0.
Otherwise, you loose the solutions of the equation cos(x) = 0.
(b) If you do it (if you divide equations (2) and (3) by cos(x) without analyzing the case cos(x) = 0),
it is traditionally considered as a RUDE MISTAKE, showing that you don't know the rules of solution of such equations.
So, doing in this way and making this error, you INEVITABLY loose your scores at an exam.