SOLUTION: find all solutions in the interval [0,2pi) for the equation 3csc x-4sin x=0.

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Question 788739: find all solutions in the interval [0,2pi) for the equation 3csc x-4sin x=0.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
csc%28x%29=1%2Fsin%28x%29does not exist when sin%28x%29=0 at x=0 and x=pi
For any other values of x, we can transform the equation into an equivalent quadratic equation by multiplying times sin%28x%29:
3csc%28x%29-4sin%28x%29=0<-->3%2Fsin%28x%29-4sin%28x%29=0-->3-4%28sin%28x%29%29%5E2=0
3-4%28sin%28x%29%29%5E2=0-->3=4%28sin%28x%29%29%5E2=0-->%28sin%28x%29%29%5E2=0=+3%2F4
The solutions come from
sin%28x%29=sqrt%283%2F4%29=sqrt%283%29%2F2 and
sin%28x%29=-sqrt%283%2F4%29=-sqrt%283%29%2F2
In the interval %22%5B0%2C%222pi%22%29%22
sin%28x%29=sqrt%283%29%2F2-->system%28highlight%28x=pi%2F3%29%2C+%22or%22%2Chighlight%28x=2pi%2F3%29%29 ( sin%28pi%2F3%29=sin%2860%5Eo%29=sqrt%283%29%2F2 and being a supplementary angle 2pi%2F3=pi-pi%2F3 has the same sine)
and sin%28x%29=-sqrt%283%29%2F2-->system%28highlight%28x=4pi%2F6%29%2C+%22or%22%2Chighlight%28x=5pi%2F6%29%29 (Those are co-terminal angles of -pi%2F3 and -2pi%2F3, found adding 2pi to -pi%2F3 and -2pi%2F3 )

NOTE:
For this kind of equation, a change of variable often helps to see the equation as a quadratic equation.
It is easier to see 3-4%28sin%28x%29%29%5E2=0 as a quadratic equation using the change of variable sin%28x%29=y.
In this case, it was easy to solve the problem, even without mentioning a quadratic equation. In other cases, solving the quadratic may be more complivcated, so making the change of variable may save confusion and ink.
We get 3-4y%5E2=0