# Solve trigonometry problem

1. Dec 4, 2016

### chwala

1. The problem statement, all variables and given/known data

Solve $2sin3x.sinx=1$

2. Relevant equations

3. The attempt at a solution
I used the identity $( cos (A+B)= cos A cos B- sin A sin B), (cos(A-B) = cos A cos B+sinA sinB)→ -(cos 4x-cos2x)= 2sin 3xsinx, (cos 2x-cos4x=1)$

now i am stuck , is this correct_
using $(cos 2x=2cos^2x-1)⇒(cos 4x=2cos^22x-1),→(2cos^22x-2cos^2x+1=0)$

is this correct, how do i move folks

2. Dec 4, 2016

### cnh1995

I'd suggest you use the triple angle formula for sine. It will get much simpler.
Edit: No I think your current approach is good. Solve the quadratic equation using the discriminant method.

3. Dec 4, 2016

### ehild

Substitute 2cos2(2x)-1 for cos(4x) into the equation cos(2x)-cos(4x)=1, and solve for cos(2x) .

4. Dec 4, 2016

### chwala

Are you suggesting i use $sin 3x≡3sin x- 4sin^3x$

5. Dec 4, 2016

### chwala

ehild hahahahahahahha that is what i was looking for lol, greetings from Africa bro

6. Dec 4, 2016

### cnh1995

No.. I edited my previous post. I was talking about the approach in #3, but it looks like I misread your quadratic equation.

7. Dec 4, 2016

### chwala

in that case i just have to solve $cos 2x(2cos 2x-1)=0$ thanks folks

8. Dec 4, 2016

### chwala

what do you mean by saying solve the quadratic using discriminant method....

9. Dec 4, 2016

### ehild

grandma. :) You are welcome.

10. Dec 4, 2016

### cnh1995

11. Dec 5, 2016

### ehild

The name is "quadratic formula" . Discriminant method is something different.