Solve the equation.
sin (x + 
) =0
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We know that principal value branch of
![{sin}^{ - 1} x = [- \frac{\pi}{2} \frac{\pi}{2}] \\](https://tex.z-dn.net/?f=+%7Bsin%7D%5E%7B+-+1%7D+x+%3D+%5B-+%5Cfrac%7B%5Cpi%7D%7B2%7D+%5Cfrac%7B%5Cpi%7D%7B2%7D%5D+%5C%5C+ "{sin}^{ - 1} x = [- \frac{\pi}{2} \frac{\pi}{2}] \\")
To solve the given equation we must keep consider that sin and sin inverse cancels each other only if x belongs to principal value .
![sin (x +\frac{\pi }{5}) =0 \\ \\ (x +\frac{\pi }{5}) = {sin}^{ - 1} (0) \\ \\ we \: know \: that \: sin \: 0° = 0 \\ \\ (x +\frac{\pi }{5}) = {sin}^{ - 1} (sin \: 0°) \\ \\ (x +\frac{\pi }{5}) = 0 \: \: \: \: (x \: belongs \: to \: [- \frac{\pi}{2} \frac{\pi}{2}]) \\ \\ x = - \frac{\pi}{5} \\ \\](https://tex.z-dn.net/?f=sin+%28x+%2B%5Cfrac%7B%5Cpi+%7D%7B5%7D%29+%3D0+%5C%5C+%5C%5C+%28x+%2B%5Cfrac%7B%5Cpi+%7D%7B5%7D%29+%3D+%7Bsin%7D%5E%7B+-+1%7D+%280%29+%5C%5C+%5C%5C+we+%5C%3A+know+%5C%3A+that+%5C%3A+sin+%5C%3A+0%C2%B0+%3D+0+%5C%5C+%5C%5C+%28x+%2B%5Cfrac%7B%5Cpi+%7D%7B5%7D%29+%3D+%7Bsin%7D%5E%7B+-+1%7D+%28sin+%5C%3A+0%C2%B0%29+%5C%5C+%5C%5C+%28x+%2B%5Cfrac%7B%5Cpi+%7D%7B5%7D%29+%3D+0+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%28x+%5C%3A+belongs+%5C%3A+to+%5C%3A+%5B-+%5Cfrac%7B%5Cpi%7D%7B2%7D+%5Cfrac%7B%5Cpi%7D%7B2%7D%5D%29+%5C%5C+%5C%5C+x+%3D+-+%5Cfrac%7B%5Cpi%7D%7B5%7D+%5C%5C+%5C%5C+ "sin (x +\frac{\pi }{5}) =0 \\ \\ (x +\frac{\pi }{5}) = {sin}^{ - 1} (0) \\ \\ we \: know \: that \: sin \: 0° = 0 \\ \\ (x +\frac{\pi }{5}) = {sin}^{ - 1} (sin \: 0°) \\ \\ (x +\frac{\pi }{5}) = 0 \: \: \: \: (x \: belongs \: to \: [- \frac{\pi}{2} \frac{\pi}{2}]) \\ \\ x = - \frac{\pi}{5} \\ \\")
is the solution of the equation.
To solve the given equation we must keep consider that sin and sin inverse cancels each other only if x belongs to principal value .
is the solution of the equation.
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