Hello guys this is my first time posting here.I need help to understand the question and the answer I have posted in the images.I need to know what this question is asking and how is the question being solved.I know trignometric functions but do not know how to solve these type of trignometric questions.I will be solving these types of questions in my exams.I am waiting for your answers so I can learn.Thanks!
For a problem like this, I like to make a quick substitution: $x = \cos\theta$. Then,
$$2 \cos^2 \theta - \cos\theta = 0 \;\;\; \text{becomes} \;\;\; 2x^2 - x = 0$$
Solve for $x$ in this latter equation to get some solutions.
So say your solutions were hypothetically $x = 1$ and $x=0$ (note: these are hypothetical solutions to an equation, not necessarily solutions to the one above). Then, remembering $x = \cos \theta$,
$$x = 1 \;\;\; \text{means} \;\;\; \cos \theta = 1$$ $$x = 0 \;\;\; \text{means} \;\;\; \cos \theta = 0$$
From there, you want to find which $\theta$ in the interval in the problem yield the values you found. (Don't forget that $\cos\theta$ and other related functions are periodic, which may prove helpful sometimes. So if $\cos (\theta) =1$, so does $\cos (\theta + 2\pi)$, for example.)
Plugging in the values for $\theta$ you find into the original equation will also prove helpful to double-check them. (I forget if this method introduces extraneous solutions. I don't think it does but I could be wrong. Regardless, checking solutions like this is always a good paranoia check.)
If none of the values in the given interval yield a solution to the equation, then you say the solution set is $\emptyset$: this means "this solution set is empty," or equivalently "there are no solutions."