Solce for x: sin-1x=cos-1x a question from Inverse Trigonometry Functions
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sin⁻¹x=cos⁻¹x
or, sin⁻¹x=π/2-sin⁻¹x [∵, sin⁻¹x+cos⁻¹x=π/2]
or, 2sin⁻¹x=π/2
or, sin⁻¹x=π/4
or, x=sin(π/4)
or, x=1/√2
or, sin⁻¹x=π/2-sin⁻¹x [∵, sin⁻¹x+cos⁻¹x=π/2]
or, 2sin⁻¹x=π/2
or, sin⁻¹x=π/4
or, x=sin(π/4)
or, x=1/√2
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