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Hi can anyone help me solve with the following equations please?

a) sinx=3cosx

b) 6sin2x=5cosx+7

c) sin2x-2=2cos2x-4sinx

a) sinx=3cosx

b) 6sin2x=5cosx+7

c) sin2x-2=2cos2x-4sinx

for part a) divide both sides by cosx. this will give you

tan x= 3

x= arctan 3

tan x= 3

x= arctan 3

part b) use the identity sin2x= 2sinxcosx

12sinxcosx= 3cosx

rewrite- 12sinxcosx- 3cosx= 0

factorise= 3cosx(4sinx- 1) =0

so 3cosx=0 and 4sinx-1 =0

12sinxcosx= 3cosx

rewrite- 12sinxcosx- 3cosx= 0

factorise= 3cosx(4sinx- 1) =0

so 3cosx=0 and 4sinx-1 =0

Thank you

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