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  4. The roots of the equation 6 sin -1( x3-6x2+8x+(1/2) )=π , are

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Q. The roots of the equation $ 6\,{{\sin }^{-1}}\left( {{x}^{3}}-6{{x}^{2}}+8x+\frac{1}{2} \right)=\pi , $ are

J & K CETJ & K CET 2015Inverse Trigonometric Functions

Solution:

Given equations is $ 6{{\sin }^{-1}}\left( {{x}^{3}}-6{{x}^{2}}+8x+\frac{1}{2} \right)=\pi $
$ \Rightarrow $ $ {{\sin }^{-1}}\left( {{x}^{3}}-6{{x}^{2}}+8x+\frac{1}{2} \right)=\frac{\pi }{6} $
$ \Rightarrow $ $ {{x}^{3}}-6{{x}^{2}}+8x+\frac{1}{2}=\sin \frac{\pi }{6} $
$ \Rightarrow $ $ {{x}^{3}}-6{{x}^{2}}+8x+\frac{1}{2}=\frac{1}{2} $
$ \Rightarrow $ $ x({{x}^{2}}-6x+8)=0 $
$ \Rightarrow $ $ x(x-2)(x-4)=0 $
So, roots are $ x=0,2,4 $
Hence, these roots are in AP.