Question

If \[\alpha \] is the root of the equation \[25{\cos ^2}\theta + 5\cos \theta - 12 = 0\] , \[\dfrac{\pi }{2} < \alpha < \pi \] then \[\sin 2\alpha \] is equals to
A. \[\dfrac{{24}}{{25}}\]
B. \[\dfrac{{ - 24}}{{25}}\]
C. \[\dfrac{{13}}{{18}}\]
D. \[\dfrac{{ - 13}}{{18}}\]

Answer 1

Hint: This is simply based on the knowledge of quadratic equations. To find the value of \[\sin 2\alpha \] we need to find the value of \[\alpha \]that is the root of the equation. For that there are various methods. But here we will use breaking the middle term. After that we will get two roots from which depending on the condition given for \[\alpha \] we will decide the root. Then the value of \[\sin 2\alpha \].

Complete step by step answer:
Given the equation is,
\[25{\cos ^2}\theta + 5\cos \theta - 12 = 0\]
Now on splitting the middle term,
\[25{\cos ^2}\theta + 20\cos \theta - 15\cos \theta - 12 = 0\]
Now taking \[5\cos \theta \] common from first two terms and -3 common from last two terms,
\[5\cos \theta \left( {5\cos \theta + 4} \right) - 3\left( {5\cos \theta + 4} \right) = 0\]
Separate the brackets,
\[\left( {5\cos \theta + 4} \right)\left( {5\cos \theta - 3} \right) = 0\]
Equating the brackets separately to zero we get,
\[\cos \theta = \dfrac{{ - 4}}{5},\dfrac{3}{5}\]
Now here \[\theta = \alpha \] thus we can say that \[\cos \alpha = \dfrac{{ - 4}}{5},\dfrac{3}{5}\]
But the condition is, \[\dfrac{\pi }{2} < \alpha < \pi \] that is it lies in the second quadrant. And as we know that in the second quadrant cos function is negative. So
\[\cos \alpha = \dfrac{{ - 4}}{5}\]
Now we know that,
\[\sin 2\alpha = 2\sin \alpha \cos \alpha \]
Thus lets find the value for \[\sin \alpha \]
\[\sin \theta = \pm \sqrt {1 - {{\cos }^2}\theta } \]
Putting the value for \[\cos \alpha = \dfrac{{ - 4}}{5}\]
\[\sin \theta = \pm \sqrt {1 - \dfrac{{16}}{{25}}} \]
Taking the LCM,
\[\sin \theta = \pm \sqrt {\dfrac{{25 - 16}}{{25}}} \]
\[\sin \theta = \pm \sqrt {\dfrac{9}{{25}}} \]
Taking the root,
\[\sin \theta = \pm \dfrac{3}{5}\]
Since \[\theta = \alpha \] we can write, \[\sin \alpha = \pm \dfrac{3}{5}\]
But \[\dfrac{\pi }{2} < \alpha < \pi \] an din second quadrant sin is positive. Thus, \[\sin \alpha = \dfrac{3}{5}\]
Now we have value of both sin and cos, putting them in the formula above we can write,
\[
  \sin 2\alpha = 2\sin \alpha \cos \alpha \\
  \sin 2\alpha = 2 \times \dfrac{3}{5} \times \dfrac{{ - 4}}{5} \\
\]
On multiplying we get,
\[\sin 2\alpha = \dfrac{{ - 24}}{{25}}\]

So, the correct answer is “Option B”.

Note: Here don’t get confused in \[\theta \& \alpha \]. Both are the same, only used in different places. But the quadrant restriction is also applicable for both. Note that the ASTC rule is important. Because that is the only rule that helped us to find the exact which value to be further used.
ASTC stands for All, Sin, Tan and Cos. These are positive in the first, second, third and fourth quadrant respectively.