Question

The number of solutions 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 can equate the sine function to a variable to substitute in the given equation to get a quadratic equation. Then we can solve the quadratic equation for the new variable using the method of factorization. From this, we can find the principal solution of the sine function. Using the principle value, we can find the solutions in the given interval and get the number of solutions.

Complete step by step Answer:

We have the equation $2{\sin ^2}x + 5\sin x - 3 = 0$
Let ${\text{y = sinx}}$. We can make this substitution in the above equation. We get,
$2{y^2} + 5y - 3 = 0$
We can solve the equation using the method of factorizing.
$
  2{y^2} + 5y - 3 = 0 \\
   \Rightarrow 2{y^2} + 6y - 1y - 3 = 0 \\
 $
On taking common factors, we get,
$
   \Rightarrow 2y\left( {y + 3} \right) - 1\left( {y + 3} \right) = 0 \\
   \Rightarrow \left( {2y - 1} \right)\left( {y + 3} \right) = 0 \\
 $
\[ \Rightarrow y = \dfrac{1}{2}, - 3\]
We cannot take ${\text{y = - 3}}$ as a solution as it does not lie range of ${\text{sinx}}$. So, we can take $y = \dfrac{1}{2}$as the solution. Giving back in the substitution we made, we get,
$\dfrac{1}{2} = \sin x$
$ \Rightarrow x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}$
The value of ${\text{sinx}}$ is positive only in 1st and 2nd quadrants. Therefore, in the interval $\left[ {{\text{0,3$\pi$ }}} \right]$, x can take the values,
\[
  x = \dfrac{\pi }{6} \\
  x = \pi - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} \\
 \]
\[
  x = 2\pi + \dfrac{\pi }{6} = \dfrac{{13\pi }}{6} \\
  x = 3\pi - \dfrac{\pi }{6} = \dfrac{{17\pi }}{6} \\
 \]
Therefore, x can have 4 values in the interval $\left[ {{\text{0,3$\pi$ }}} \right]$ that satisfies the given equation.
So, the correct answer is option C.

Note: Range of the sine function is the interval $\left[ {{\text{ - 1,1}}} \right]$. So, values outside the interval can be rejected. The value of sin function is positive only in the 1st and 2nd quadrants, so we take only the values in 1st and 2nd quadrants. We must understand the method of finding the solutions in other quadrants using the principle solution. Solving the quadratic equation can be done by either factorizing or using the quadratic formula.