sinophile said:
Solve for 0<=x<=360:
2sin^2(theta)+cos^2(theta)=2
I can't seem to isolate this equation into only one type of ratio. Am I supposed to isolate it to one ratio? What the hell do I do?
Notice how 2sin
2(theta) + cos
2(theta) is like sin
2(theta) + cos
2(theta)?
And we know that sin
2(theta) + cos
2(theta) equals 1.
So we can try to work around this, factoring wont works, but if you see
2sin
2(theta) + cos
2(theta) = 1 + 1
2sin
2(theta) = 1 + 1 - cos
2(theta)
2sin
2(theta) - 1 = 1 - cos
2(theta)
And from rearranging sin
2(theta) + cos
2(theta) = 1, we have
sin
2(theta) = 1 - cos
2(theta)
this is the RHS of the equation 2sin
2(theta) - 1 = 1 - cos
2(theta),
So making the substitution,
2sin
2(theta) - 1 = sin
2(theta)
2sin
2(theta) - sin
2(theta) = 1
sin
2(theta) = 1
therefore
(sin(theta))
2 = 1
sin(theta) = +1 and -1 (from the root)
So then you just have to solve that equation for 0<= x <= 360.
If you need any help, just ask.
Goodluck!