Find all solutions in the interval [0, 2π). 2 sin2x = sin x x = pi divided by three. , two pi divided by three. x = pi divided by two. , three pi divided by two. , pi divided by three. , two pi divided by three. x = 0, π, pi divided by six , five pi divided by six x = pi divided by six , five pi divided by six

All solutions of the equation are .

Step-by-step explanation:

I am assuming that the equation is .

Notice that the equation can be written as . 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

Hence, the equation becomes

Extracting the common factor:

Then, we need to solve two simpler equations:

and

Let us start with the first one. The sinus is zero in 0, and , but this last point isn't included in the interval. Then, all the solutions in are {0, }.

In the second equation we have , so . Now, notice that this solution is in the first quadrant, and we are looking for all the solutions in , then we need to include the co-terminal angle in the fourth quadrant. Using the reduction formulas, the other solution is .

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