Can't you say
Okay, the equation is sin 4x = -2 sin 2x. I manipulate and rearrange the equation to acquire 2 sin 2x(cos 2x + 1) = 0. I set each factor to zero, but come across a problem when setting the factor 2 sin 2x to zero. How I did, which was different from the answer key, was take the angle that makes sin 2x = 0, which was 0, and add 2npi.
So: 2x = 0 + 2npi ---> x = npi. But they have x = (n/2)pi, giving them two more angles than I had produced--which were 0 and pi. Why did they do that? They want me to find all of the angles on the interval [0, 2pi)
I understandly perfectly what you are saying. It is just sort of contentious when comparing it to the way my professor taught me. She would tell me that if sin 2x = 0
then 2x = 0 + 2npi ----> x = (0 + 2npi)/2 ------> x =npi. Then, I would continue to input values in for n, until I had all of the angles in the interval [0, 2pi) That's why I don't understand why it would be (n/2)pi.
Then you have misunderstood what your professor has said. I have explained why it is , arguing about it won't make what you think you heard correct.
What she would have said is that when you solve a trigonometric equation, you need to find the solution from the TWO QUADRANTS that satisfy the equation, then add multiples of .
IF this is the case, then in this special case, where , you would find that , and when you write down the first few values, you'll see that the pattern gives . You are incorrect because each time you have forgotten about the solution and all of the odd integer multiples.