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SSC CHSL 2021 PYQ — Question 64

SSC CHSL - 15 April 2021 - Shift 2 • Quantitative Aptitude • Q64

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Solve for x:

sin2 x - 4 sin x + 3 = 0, 0 ≤\leq x ≤\leq π2\frac{\pi}{2}

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Correct answer: C

Explanation

The correct answer is option 3 i.e.  π2\frac{\pi}{2}

 sin2 x - 4 sin x + 3 = 0

=>sin2 x - 3 sin x-sin x + 3 = 0

=>sin x(sin x - 3)-1(sin x - 3)=0

=>(sin x - 1)(sin x - 3)=0

=>sin x - 1 =0                    [∵ sin x can't be equals to 3]

=> sin x = 1

∵ sin π2\frac{\pi}{2}= 1

=>x= π2\frac{\pi}{2}