Find the general solutions for 3sinx+4cosx=5

Steps to solve Find the general solutions for 3sinx+4cosx=5

⇒3sinx+4cosx=5

Put 3=rcosθ and 4=rsinθ

⇒3^2.4^2=r^2cos^2θ+r^2sin^2θ

⇒25=r^2(cos^2θ+sin^2θ)

⇒r^2=25

⇒r=±5

tanθ=rsinθ/rcosθ=4/3

⇒θ=tan^(-1)(4/3)

⇒3sinx+4cosx=5

⇒rcosθ.sinx+rsinθ.cosx=5

⇒sin(x+θ)=5/5

⇒r+θ=nπ+(-1)^n(π/2)

⇒x=nπ+(-1)^n(π/2)-tan^(-1)(4/3)

Hence the general solutions for 3sinx+4cosx=5 is nπ+(-1)^n(π/2)-tan^(-1)(4/3) 

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