How do you find all solutions of the equations $2sinx+cscx=0$ in the interval $[0,2pi)$ ?

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Answer 1 · 2017-02-05T16:32:48

No solution

$2sinx+cscx=0$ can be written as

$2sinx+1/sinx=0$ and multiplying by $sinx$ we get

$2sin^2x+1=0$

But LHS can never be $0$ ,

hence we donot have any solution.

This can also be checked from the graph of $2sinx+cscx$ , which never has a value of $0$ .
graph{2sinx+cscx [-10, 10, -5, 5]}

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