Question

Solve the trigonometric expression to find the value of $x$ :
$2{{\cos }^{2}}x+3\sin x=0$.

Answer 1

Hint: Change cosine of the given angle into its sine form by using the formula: ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$. Form a quadratic equation with sine as a variable and use the middle term split method to write the equation as a product of two terms. Substitute each term equal to 0 to find the general solution of the equation. Use the relation: if $\sin A=\sin B$, then $A=n\pi +{{\left( -1 \right)}^{n}}B$, where ‘n’ is any integer.

Complete step-by-step answer:

We have been given: $2{{\cos }^{2}}x+3\sin x=0$.
We know that,
$\begin{aligned} & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\\ & \Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\\ \end{aligned}$
Therefore, substituting this value in the given equation, we get,
$\begin{aligned} & 2\left( 1-{{\sin }^{2}}x \right)+3\sin x=0 \\\ & \Rightarrow 2-2{{\sin }^{2}}x+3\sin x=0 \\\ & \Rightarrow 2{{\sin }^{2}}x-3\sin x-2=0 \\\ \end{aligned}$
Splitting the middle term, we get,
$2{{\sin }^{2}}x-4\sin x+\sin x-2=0$
Taking $2\sin x$ common from the first and second term and also taking 1 common from the third and fourth term, we get,
$2\sin x\left( \sin x-2 \right)+1\left( \sin x-2 \right)=0$
Now, taking $\left( \sin x-2 \right)$ common from the two terms, we have,
$\left( \sin x-2 \right)\left( 2\sin x+1 \right)=0$
Substituting each term equal to 0, we get,

& \sin x-2=0\text{ or 2}\sin x+1=0 \\\ & \Rightarrow \sin x=2......................(i) \\\ & \text{or }\sin x=\dfrac{-1}{2}...................(ii) \\\ \end{aligned}$$ Case (i) is not possible because the range of sine function is -1 to 1 and hence it cannot be 2. Therefore, only case (ii) is possible. $\Rightarrow \sin x=\dfrac{-1}{2}$ We know that, $\sin \left( \dfrac{-\pi }{6} \right)=\dfrac{-1}{2}$. So the above equation can be written as: $\sin x=\sin \left( \dfrac{-\pi }{6} \right)$ Now, using the general solution formula for sine function, given by: if $\sin A=\sin B$, then $A=n\pi +{{\left( -1 \right)}^{n}}B$, we have, $x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{-\pi }{6} \right)$, where ‘n’ is any integer. Note: One may note that we have found the general solution of the given equation because we haven’t been provided with any particular limit of ‘x’ in which we have to find the solution, so that means we have to generalize the solution. By putting the integral values of ‘n’ we can find an infinite number of solutions of the given equation. The basic thing in the above question is that, to find the general solution, we have to change different trigonometric functions into a single trigonometric function by applying different formulas.