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 :

"if it is my birthday, then it is snowing"

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it