Find all solutions in the interval [0, 2π).
2 sin2x = sin x

The solutions on the given interval are :

Step-by-step explanation:

We will need the double angle identity .

Let's begin:

Use double angle identity mentioned on left hand side:

Simplify a little bit on left side:

Subtract on both sides:

Factor left hand side:

Set both factors equal to 0 because at least of them has to be 0 in order for the equation to be true:

The first is easy what angles are -coordinates on the unit circle 0. That happens at and on the given range of (this is not be confused with the -coordinate).

Now let's look at the second equation:

Isolate .

Add 1 on both sides:

Divide both sides by 4:

This is not as easy as finding on the unit circle.

We know will render us a value between and .

So one solution on the given interval for x is .

We know cosine function is even.

So an equivalent equation is:

Apply to both sides:

Multiply both sides by -1:

This going to be negative in the 4th quadrant but if we wrap around the unit circle, , we will get an answer between and .

So the solutions on the given interval are :

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gimmee one of those gold crowns!

~coco

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: