Solve the Equation exactly over the interval [0, 2pi]
the problems i do not understand are:
2sin^2x + sin x - 1 = 0
2sinx cosx + cos(x) = 0
2sin^2x - sin x = 0
if anyone could help me i would appreciate it alot. thanks
Solve the Equation exactly over the interval [0, 2pi]
the problems i do not understand are:
2sin^2x + sin x - 1 = 0
2sinx cosx + cos(x) = 0
2sin^2x - sin x = 0
if anyone could help me i would appreciate it alot. thanks
Unless I'm missing something equation 4 is the same as equation 1 so you've solved that too.
Number 3:
$\displaystyle 2sin^2x - sin x = 0$
You can factor out sin(x) since it's a common factor: $\displaystyle \sin(x)(2\sin(x)-1) = 0$
Either $\displaystyle \sin(x) = 0$ or $\displaystyle 2\sin(x) -1 = 0$
Can you find x in those cases?
No. $\displaystyle \sin(x) = \dfrac{1}{2}$ but not x.
It's probably easier to start with when $\displaystyle \sin(x) = 0$ and then $\displaystyle x = 0 , \pi , 2\pi$ - delete as appropriate depending on your boundary conditions
For $\displaystyle \sin(x) = \dfrac{1}{2}$ we have $\displaystyle x = \arcsin(0.5) = \dfrac{\pi}{6}$ as the principal solution but there is another one to find