Question

The number of solution of $x$ in the interval $\left[ {0,3\pi } \right]$ satisfying the equation \[2{\sin ^2}x + 5\sin x - 3 = 0\] is
A. 1
B. 2
C. 4
D. 6

Answer 1

Hint:
We will let $\sin x = y$, then form an equation. Then, we will solve the equation by factoring the equation. We can then find the principal solution of the sine function from the solution of $x$.

Complete step by step solution:
We have to find the number of solutions of \[2{\sin ^2}x + 5\sin x - 3 = 0\]
Let $\sin x = y$
Then, the given equation can be written as $2{y^2} + 5y - 3 = 0$
Now, we will factorise the equation.
$
  2{y^2} + 6y - y - 3 = 0 \\
   \Rightarrow 2y\left( {y + 3} \right) - 1\left( {y + 3} \right) = 0 \\
   \Rightarrow \left( {2y - 1} \right)\left( {y + 3} \right) = 0 \\
$
Put each factor equal to 0 to find the value of $y$
$
  2y - 1 = 0 \\
   \Rightarrow y = \dfrac{1}{2} \\
$
And
$y = - 3$
But, $y = \sin x$ and value of $\sin x$ cannot be less than $ - 1$
Therefore, value of $y = \sin x = \dfrac{1}{2}$
Now, we know that $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$
Hence, $\sin x = \sin \dfrac{\pi }{6}$
Now, $\sin \theta $ is positive only in the first and second quadrant.
From $\left[ {0,2\pi } \right]$, there are only values of $x$ for which the equation satisfies.
And there are two more values of $\left[ {\pi ,3\pi } \right]$ which lies in the first and second quadrant.
Hence, there are 4 values in the interval $\left[ {0,3\pi } \right]$ which satisfies the given condition.

Thus, option C is correct.

Note:
Students must know the range of the sine function. The range of the sine function is $\left[ { - 1,1} \right]$. Then, values of function outside the interval should not be taken. We have solved the quadratic equation using factorisation, but we can solve it using the quadratic formula.