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Question: The number of values of \[\alpha \] in \[\left[ {0,2\pi } \right]\] for which \[{\text{2si}}{{\text{...

The number of values of α\alpha in [0,2π]\left[ {0,2\pi } \right] for which 2sin3α - 7sin2α + 7sinα=2{\text{2si}}{{\text{n}}^{\text{3}}}\alpha {\text{ - 7si}}{{\text{n}}^{\text{2}}}\alpha {\text{ + 7sin}}\alpha = {\text{2}}, is:
A) 6
B) 4
C) 3
D) 1

Explanation

Solution

Here we will use the identity:
a3−b3=(a−b)(a2+b2+ab){a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right) and then simplify the resultant term and then find the values of α\alpha accordingly.

Complete step-by-step answer:
The give equation is:-
2sin3α - 7sin2α + 7sinα=2{\text{2si}}{{\text{n}}^{\text{3}}}\alpha {\text{ - 7si}}{{\text{n}}^{\text{2}}}\alpha {\text{ + 7sin}}\alpha = {\text{2}}
Simplifying it further we get:-
2sin3α - 7sin2α + 7sinα−2 = 0{\text{2si}}{{\text{n}}^{\text{3}}}\alpha {\text{ - 7si}}{{\text{n}}^{\text{2}}}\alpha {\text{ + 7sin}}\alpha - {\text{2 = 0}}
Taking out the terms as common we get:-
2(sin3α−1)−7sinα(sin⁡α−1)=02\left( {{\text{si}}{{\text{n}}^{\text{3}}}\alpha - 1} \right) - 7{\text{sin}}\alpha \left( {\sin \alpha - 1} \right) = 0
Now applying the following identity:
a3−b3=(a−b)(a2+b2+ab){a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)
We get:-
2(sin⁡α−1)(sin⁡2α+1+(1)sin⁡α)−7sinα(sin⁡α−1)=02\left( {\sin \alpha - 1} \right)\left( {{{\sin }^2}\alpha + 1 + \left( 1 \right)\sin \alpha } \right) - 7{\text{sin}}\alpha \left( {\sin \alpha - 1} \right) = 0
Now simplifying it further we get:-
2(sin⁡α−1)(sin⁡2α+1+sin⁡α)−7sinα(sin⁡α−1)=02\left( {\sin \alpha - 1} \right)\left( {{{\sin }^2}\alpha + 1 + \sin \alpha } \right) - 7{\text{sin}}\alpha \left( {\sin \alpha - 1} \right) = 0
Now taking (sin⁡α−1)\left( {\sin \alpha - 1} \right) as common we get:-
Solving it further we get:-

(sin⁡α−1)[2sin⁡2α+2+2sin⁡α−7sinα]=0 (sin⁡α−1)[2sin⁡2α+2−5sin⁡α]=0  \left( {\sin \alpha - 1} \right)\left[ {2{{\sin }^2}\alpha + 2 + 2\sin \alpha - 7{\text{sin}}\alpha } \right] = 0 \\\ \left( {\sin \alpha - 1} \right)\left[ {2{{\sin }^2}\alpha + 2 - 5\sin \alpha } \right] = 0 \\\

Now solving the quadratic equation using middle term split we get:-
(sin⁡α−1)[2sin⁡2α−4sin⁡α−sin⁡α+2]=0\left( {\sin \alpha - 1} \right)\left[ {2{{\sin }^2}\alpha - 4\sin \alpha - \sin \alpha + 2} \right] = 0
Solving it further we get:-
⇒(sin⁡α−1)(2sin⁡α−1)(sin⁡α−2)=0\Rightarrow \left( {\sin \alpha - 1} \right)\left( {2\sin \alpha - 1} \right)\left( {\sin \alpha - 2} \right) = 0
Now evaluating the value of sin⁡α\sin \alpha we get:-

sin⁡α=1;2sin⁡α=1;sin⁡α=2 ⇒sin⁡α=1;sin⁡α=12;sin⁡α=2  \sin \alpha = 1;2\sin \alpha = 1;\sin \alpha = 2 \\\ \Rightarrow \sin \alpha = 1;\sin \alpha = \dfrac{1}{2};\sin \alpha = 2 \\\

Now since we know that −1⩽sin⁡θ⩽1 - 1 \leqslant \sin \theta \leqslant 1
Therefore, sin⁡α≠2\sin \alpha \ne 2
Therefore,
sin⁡α=1;sin⁡α=12\sin \alpha = 1;\sin \alpha = \dfrac{1}{2}
Since α\alpha is in [0,2π]\left[ {0,2\pi } \right]
Now we know that,

sin⁡π2=1 sin⁡π6,sin⁡5π6=12  \sin \dfrac{\pi }{2} = 1 \\\ \sin \dfrac{\pi }{6},\sin \dfrac{{5\pi }}{6} = \dfrac{1}{2} \\\

Hence values of α\alpha are: - π2,π6,5π6\dfrac{\pi }{2},\dfrac{\pi }{6},\dfrac{{5\pi }}{6}
Hence there are 3 values of α\alpha
Hence option C is the correct option.

Note: Students should note that sine function is positive in 1st and 2nd quadrant.

Also, the identity and the calculations should be correct and accurate.