Question

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

Answer 1

Hint: Here we will use the identity:
\[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\] and then simplify the resultant term and then find the values of \[\alpha \] accordingly.

Complete step-by-step answer:
The give equation is:-
\[{\text{2si}}{{\text{n}}^{\text{3}}}\alpha {\text{ - 7si}}{{\text{n}}^{\text{2}}}\alpha {\text{ + 7sin}}\alpha = {\text{2}}\]
Simplifying it further we get:-
\[{\text{2si}}{{\text{n}}^{\text{3}}}\alpha {\text{ - 7si}}{{\text{n}}^{\text{2}}}\alpha {\text{ + 7sin}}\alpha - {\text{2 = 0}}\]
Taking out the terms as common we get:-
\[2\left( {{\text{si}}{{\text{n}}^{\text{3}}}\alpha - 1} \right) - 7{\text{sin}}\alpha \left( {\sin \alpha - 1} \right) = 0\]
Now applying the following identity:
\[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\]
We get:-
\[2\left( {\sin \alpha - 1} \right)\left( {{{\sin }^2}\alpha + 1 + \left( 1 \right)\sin \alpha } \right) - 7{\text{sin}}\alpha \left( {\sin \alpha - 1} \right) = 0\]
Now simplifying it further we get:-
\[2\left( {\sin \alpha - 1} \right)\left( {{{\sin }^2}\alpha + 1 + \sin \alpha } \right) - 7{\text{sin}}\alpha \left( {\sin \alpha - 1} \right) = 0\]
Now taking \[\left( {\sin \alpha - 1} \right)\] as common we get:-
Solving it further we get:-
\[
  \left( {\sin \alpha - 1} \right)\left[ {2{{\sin }^2}\alpha + 2 + 2\sin \alpha - 7{\text{sin}}\alpha } \right] = 0 \\
  \left( {\sin \alpha - 1} \right)\left[ {2{{\sin }^2}\alpha + 2 - 5\sin \alpha } \right] = 0 \\
 \]
Now solving the quadratic equation using middle term split we get:-
\[\left( {\sin \alpha - 1} \right)\left[ {2{{\sin }^2}\alpha - 4\sin \alpha - \sin \alpha + 2} \right] = 0\]
Solving it further we get:-
\[ \Rightarrow \left( {\sin \alpha - 1} \right)\left( {2\sin \alpha - 1} \right)\left( {\sin \alpha - 2} \right) = 0\]
Now evaluating the value of \[\sin \alpha \]we get:-
\[
  \sin \alpha = 1;2\sin \alpha = 1;\sin \alpha = 2 \\
   \Rightarrow \sin \alpha = 1;\sin \alpha = \dfrac{1}{2};\sin \alpha = 2 \\
 \]
Now since we know that \[ - 1 \leqslant \sin \theta \leqslant 1\]
Therefore, \[\sin \alpha \ne 2\]
Therefore,
\[\sin \alpha = 1;\sin \alpha = \dfrac{1}{2}\]
Since \[\alpha \] is in \[\left[ {0,2\pi } \right]\]
Now we know that,
\[
  \sin \dfrac{\pi }{2} = 1 \\
  \sin \dfrac{\pi }{6},\sin \dfrac{{5\pi }}{6} = \dfrac{1}{2} \\
 \]
Hence values of \[\alpha \] are: - \[\dfrac{\pi }{2},\dfrac{\pi }{6},\dfrac{{5\pi }}{6}\]
Hence there are 3 values of \[\alpha \]
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.