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Solve: 3 tan x + cot x = 5 cosec x

Answer»

We have: 3 tan x + cot x = 5 cosec x 

\(3(\frac{sinx}{cosx})+(\frac{cosx}{sinx})=\frac{5}{sinx}\)

\(3\space\frac{sinx}{cosx}=\frac{5-cosx}{sinx}\) 

\(3sin^2x=5cosx-cos^2x\) 

\(3 (1-cos^2x)=5cosx-cos^2x\)

\(3-3cos^2x+cos^2x-5cosx=0\)

\(2cos^2x+5cosx-3=0\) 

\((2cosx-1)(cosx+3)=0\) 

\(cosx+3=0\)  or  \(cosx+3=0\) 

\(cosx=\frac{1}{2} \)  or  \(cosx=\frac{-3}{2}\) 

As range of cos x is [– 1, 1] 

⇒ ∴ \(cosx =\frac{1}{2}\) 

\(cosx=cos\frac{\pi}{3}\) 

\(x=2n\pi ±\frac{\pi}{3},n∈z\)



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