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← JEE Main 2014 · 9 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 88 of 90

JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 88

The number of values of α\alpha in [0,2π][0,2\pi] for which 2sin⁡3α−7sin⁡2α+7sin⁡α=22\sin^3\alpha-7\sin^2\alpha+7\sin\alpha=2, is :

Choose one option
Answer and worked solution

Official answer: C

With s=sin⁡αs=\sin\alpha:
2s3−7s2+7s−2=(s−1)(2s−1)(s−2)=0.2s^3-7s^2+7s-2=(s-1)(2s-1)(s-2)=0.

  • s=2s=2 is impossible.
  • s=1s=1 gives α=π2\alpha=\frac\pi2.
  • s=12s=\frac12 gives α=π6,5π6\alpha=\frac\pi6,\frac{5\pi}6.

So there are 3 values.