Question

Solvw this question!!!!!!!!!!!

Answer 1

HI THERE

HERE IS UR ANSWER....




  Given eqn. is written as −sin−16x=π2−sin−16x=π2 +sin−1(63–√x)+sin−1(63x)

Taking sinsin on both the sides we get

sin(−sin−16x)=sin(π2sin(−sin−16x)=sin(π2+sin−163–√x)+sin−163x)

We know that sin(−θ)=−sinθsin(−θ)=−sinθ

⇒−sin(sin−16x)=cos(sin−163–√x)⇒−sin(sin−16x)=cos(sin−163x)

But we know that sin−1x=cos−11−x2−−−−−√sin−1x=cos−11−x2

⇒sin−163–√x=cos−11−(63–√x)2−−−−−−−−−−√=cos−11−108x2−−−−−−−−√⇒sin−163x=cos−11−(63x)2=cos−11−108x2

and we also know that sin−1(sinθ)=θsin−1(sinθ)=θ

⇒−6x=cos(cos−11−108x2−−−−−−−−√⇒−6x=cos(cos−11−108x2

⇒−6x=1−108x2−−−−−−−−√⇒−6x=1−108x2

Squaring on both the sides we get

36x2=1−108x236x2=1−108x2

⇒144x2=1⇒144x2=1

⇒x2=1144⇒x2=1144

⇒x=±112⇒x=±112

But x=112x=112 does not satisfy the eqn.

∴x=−112∴x=−112

HOPE HELPED

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