How do you solve $3sin(2x) = 5cosx$ ?

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Answer 1 · 2016-04-16T10:41:26

General solution for $x$ is $(2n+1)xx90^o$ or $(360n+56.44)^o$ or $(360n+123.56)^o$ , where $n$ is an integer.

$3sin2x=5cosx$ can be simplified using formula $sin2A=2sinAcosA$ . Then this becomes

$3xx2sinxcosx-5cosx=0$ or

$cosx(6sinx-5)=0$

Hence either $cosx=0$ i.e. $x=(2n+1)xx90^o$ or

$sinx=5/6$ i.e. $x=56.44^o$ or $x=123.56^o$

General Solution would be $(360n+56.44)^o$ or $(360n+123.56)^o$

How do you solve 3sin(2x) = 5cosx3sin(2x) = 5cosx?