Answer:
All solutions of the equation are [tex]{0,\pi, \arccos\frac{1}{4}, 2\pi-\arccos\frac{1}{4}}[/tex].
Step-by-step explanation:
I am assuming that the equation is [tex]2\sin 2x-\sin x = 0[/tex].
Notice that the equation can be written as [tex]2\sin 2x-\sin x = 0[/tex]. Always is better to have the equations equated to zero. The next step is to use some trigonometrical identities, in this case we will use
[tex]\sin(2x) = 2\sin x\cos x.[/tex]
Hence, the equation becomes
[tex]4\sin x\cos x-\sin x =0.[/tex]
Extracting the common factor:
[tex]\sin x(4\cos x -1)=0.[/tex]
Then, we need to solve two simpler equations:
[tex]\sin x =0[/tex] and [tex]4\cos x-1=0[/tex]
Let us start with the first one. The sinus is zero in 0, [tex]\pi[/tex] and [tex]2\pi[/tex], but this last point isn't included in the interval. Then, all the solutions in [tex][0,2\pi)[/tex] are {0, [tex]\pi[/tex]}.
In the second equation we have [tex]\cos x=\frac{1}{4}[/tex], so [tex]x=\arccos\frac{1}{4}[/tex]. Now, notice that this solution is in the first quadrant, and we are looking for all the solutions in [tex][0,2\pi)[/tex], then we need to include the co-terminal angle in the fourth quadrant. Using the reduction formulas, the other solution is [tex]2\pi-\arccos\frac{1}{4}[/tex].
