Question

# Solve the following equation for x: sin−1(1−x)−2sin−1x=π2, then x is equal to a) 0,12 b) 1,12 c) zero d) 12

Solution

## Given sin−1(1−x)−2sin−1x=π2⇒  −2sin−1x=π2−sin−1(1−x)⇒−2sin−1x=cos−1(1−x)       [∵sin−1(1−x)+cos−1(1−x)=π2] ⇒  cos(−2sin−1x)=1−x    [multiplying both sides by cosx] ⇒  cos(2sin−1x)=1−x      [∵cos(−x)=+cos x]⇒  [1−2sin2(sin−1x)]=1−x      [∵ cos2x=1−2sin2x]⇒  1−2[sin(sin−1x)]2=1−x      [∵ sin2x=(sin x)2]⇒  1−2x2=1−x⇒2x2−x=0 ⇒  x(2x−1)=0 ⇒ x=0 or 2x−1=0  ⇒ x=0 or 12 But x=12 does not satisfy the given equation, so x = 0 Hence,the correct option is (c). Alternate method. Given, sin−1(1−x)−2sin−1x=π2 Let x=sinθ⇒θ=sin−1x, then sin−1(1−sin θ)−2θ=π2⇒sin−1(1−sinθ)=π2+2θ⇒ 1−sinθ=sin(π2+2θ)⇒1−sinθ=cos2θ[∵ sin(π2+θ)=cosθ]⇒ 1−sinθ=1−2sin2θ      [∵ cos2θ=1−2sin2θ]⇒ 2sin2θ−sinθ=0⇒sinθ(2sinθ−1)=0 Either sinθ or 2sinθ−1=0 ⇒ x=0 or2x−1=0      [∴sinθ=x]⇒ x=0 or x=12 But x=12 does not satisfy the given equation.  So, x=0 To check your answer Put x=0 in the given equation, ∴ sin−1(1−0)−2sin−10=π2⇒π2−2×0=π2⇒π2=π2 Put x=12 in the given equation, ∴ sin−1(1−12)−2sin−112=π2⇒sin−212−2sin−112=π2⇒π6−2×π6=π2⇒π−2π6=π2⇒−π6≠π2,so x=12 is not possibile Note:  While solving the trigonometric equation, sometimes it may have same extra roots, so please take careful about it. Suggest corrections Similar questions View More...

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