4 sin^-1x+cos^-1x=pi then what is the value of x
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As we know sin^-1x + cos^-1x is π÷2(The proof is a bit lengthy
4 sin^-1x can be written as 3 sin inverse x + sin inverse x
Therefore 4 sin inverse x + cos inverse x can be written as 3 sin inverse x + sin inverse x +cos inverse x=π
which will be 3 sin inverse x +π÷2=π
=>3 sin inverse x = π÷2
=>sin inverse x =π÷6
![x = \sin( \frac{\pi}{6} ) x = \sin( \frac{\pi}{6} )](https://tex.z-dn.net/?f=x+%3D++%5Csin%28+%5Cfrac%7B%5Cpi%7D%7B6%7D+%29+)
x=1÷2 or x=0.5
4 sin^-1x can be written as 3 sin inverse x + sin inverse x
Therefore 4 sin inverse x + cos inverse x can be written as 3 sin inverse x + sin inverse x +cos inverse x=π
which will be 3 sin inverse x +π÷2=π
=>3 sin inverse x = π÷2
=>sin inverse x =π÷6
x=1÷2 or x=0.5
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