Question

Determine the general solution of this equation: 2sinx = 3 + 2cscx.

I *think* I have to turn this into a quadratic equation somehow, but I'm not quite sure how to go about that. I tried 2sinx-(1/sinx)-3=0, and then multiplying everything by (sinx/sinx) to get (2sin^2x-sinx-3sinx)/(sin^2x), but then the denominator becomes a problem. How can this be simplified into something workable?

Answer 1

\[2\sin x = 3 + \frac{2}{\sin x}\]
multiply all terms by sin x
\[2\sin^2 x = 3\sin x + 2\]
let a = sin x
\[2a^2 = 3a + 2\]
put everything on the left side
\[2a^2 - 3a - 2 = 0\]
factor out
\[(2a +1)(a-2) = 0\]
solve for a
\[a = -\frac 12; \; a = 2\]
change a back to sin x
\[\sin x = -\frac 12; \; \sin x = 2\]
are you supposed to solve for x?

Answer 2

Yeah you have figured out the method right...


Sin x=2 is impossible .. So, Sin x= -1/2 should give the solution as, x=210 and 330