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

1 Answer

Shwetank Mauria

Apr 16, 2016

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

Explanation:

#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#

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