Solve the equation.
cos (x +
) = 0
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As we know that principal value branch of cos inverse is [0,π]
thus while solving the equation we must remember that cos inverse cancel cos x only if x belongs to principal value branch.
we know that cos (π/2) = 0
![cos( x+\frac{\pi }{10})= 0\\ \\ ( x+\frac{\pi }{10})= {cos}^{-1}(0) \\ \\ ( x+\frac{\pi }{10}) = {cos}^{ - 1} (cos (\frac{\pi}{2})) \\ \\ cos( x+\frac{\pi }{10})= 0\\ \\ ( x+\frac{\pi }{10})= {cos}^{-1}(0) \\ \\ ( x+\frac{\pi }{10}) = {cos}^{ - 1} (cos (\frac{\pi}{2})) \\ \\](https://tex.z-dn.net/?f=cos%28+x%2B%5Cfrac%7B%5Cpi+%7D%7B10%7D%29%3D+0%5C%5C+%5C%5C+%28+x%2B%5Cfrac%7B%5Cpi+%7D%7B10%7D%29%3D+%7Bcos%7D%5E%7B-1%7D%280%29+%5C%5C+%5C%5C+%28+x%2B%5Cfrac%7B%5Cpi+%7D%7B10%7D%29+%3D+%7Bcos%7D%5E%7B+-+1%7D+%28cos+%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%29+%5C%5C+%5C%5C+)
here both cancels with each other because π/2 belongs to [0,π]
Thus
thus while solving the equation we must remember that cos inverse cancel cos x only if x belongs to principal value branch.
we know that cos (π/2) = 0
here both cancels with each other because π/2 belongs to [0,π]
Thus
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