Question

The number of solutions of the equation $$8{\tan }^2\theta +9=6\sec \theta$$ in the interval $$\left({\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$$.
A.Two
B.Four
C.Zero
D.None of these

Answer 1

Hint: Here, we have to find the number of solutions. First, we have to solve the given equation to find the number of solutions. Trigonometric equation is an equation involving one or more trigonometric ratios of unknown angles.

Formula Used:
We will use the trigonometric identities $${\tan }^2\theta ={\sec }^2\theta -1$$ and $$\sec \theta ={\displaystyle \frac{1}{\cos \theta }}$$;

Complete step-by-step answer:
We will first solve the given equation $$8{\tan }^2\theta +9=6\sec \theta$$.
By using the trigonometric identity $${\tan }^2\theta ={\sec }^2\theta -1$$, we get
$$\Rightarrow 8({\sec }^2\theta -1)+9=6\sec \theta$$
Multiplying the terms, we get
$$\Rightarrow 8{\sec }^2\theta -8+9=6\sec \theta$$
Subtracting the like terms, we get
$$\Rightarrow 8{\sec }^2\theta +1=6\sec \theta$$
Rewriting the above equation, we get
$$\Rightarrow 8{\sec }^2\theta -6\sec \theta +1=0$$
Above equation is a quadratic equation, so we will factorize the equation to find the value of $$\theta$$.
Factorizing by grouping terms, we get
$$\Rightarrow 8{\sec }^2\theta -4\sec \theta -2\sec \theta +1=0$$
Now factoring out the common term, we get
$$\Rightarrow 4\sec \theta \left(2\sec \theta -1\right)-1\left(2\sec \theta -1\right)=0$$

$$\Rightarrow \left(4\sec \theta -1\right)\left(2\sec \theta -1\right)=0$$
Using zero product property, we get
$$\begin{array}{l}\Rightarrow \left(4\sec \theta -1\right)=0\\ \Rightarrow 4\sec \theta =1\\ \Rightarrow \sec \theta ={\displaystyle \frac{1}{4}}\end{array}$$
Or
$$\begin{array}{l}\Rightarrow \left(2\sec \theta -1\right)=0\\ \Rightarrow 2\sec \theta =1\\ \Rightarrow \sec \theta ={\displaystyle \frac{1}{2}}\end{array}$$
By using trigonometric identity $$\sec \theta ={\displaystyle \frac{1}{\cos \theta }}$$, we get
$$\Rightarrow \cos \theta =4$$
Or
$$\Rightarrow \cos \theta =2$$
We know that $$\cos \theta$$ lies between $$-1<\theta <1$$.
Since the values of $$\cos \theta$$ does not lie between $$-1$$ and 1, so, there is no solution.
Therefore, the number of solutions of the equation $$8{\tan }^2\theta +9=6\sec \theta$$ in the interval $$\left({\displaystyle \frac{-\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$$ is zero.
Hence, the correct option is option C.

Note: We know that trigonometric equations are expressed as ratios of sine, cosine, tangent, cotangent, secant, cosecant angles. All possible values that satisfy the given trigonometric equation are called solutions of the given trigonometric equation. For a complete solution, “all possible values” satisfying the equation must be obtained. When we try to solve a trigonometric equation, we try to find out all sets of values of $$\theta$$, which satisfy the given equation.