Q.1. How many real solutions are there to the equation?

We are asked to find how many real solutions there are to the equation:

$$2 \cos^4 \theta - 5 \cos^2 \theta + 3 = 0$$

in the interval $0 \leq \theta \leq 2\pi$.

Answer Choices

• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8

Solution

We can treat the equation as a quadratic in terms of $\cos^2 \theta$.

Step 1: Substitution

Let $x = \cos^2 \theta$, so the equation becomes:

$$2x^2 - 5x + 3 = 0$$

Step 2: Solve the quadratic equation

We solve this quadratic equation using the quadratic formula:

$$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(3)}}{2(2)}$$

$$x = \frac{5 \pm \sqrt{25 - 24}}{4}$$

$$x = \frac{5 \pm \sqrt{1}}{4}$$

$$x = \frac{5 \pm 1}{4}$$

This gives two solutions for $x$:

$$x_1 = \frac{5 + 1}{4} = \frac{6}{4} = 1.5$$

$$x_2 = \frac{5 - 1}{4} = \frac{4}{4} = 1$$

Step 3: Analyze the values of $x$

We know that $x = \cos^2 \theta$, and $\cos^2 \theta$ is always between 0 and 1, i.e., $0 \leq \cos^2 \theta \leq 1$. Therefore, $x_1 = 1.5$ is not valid, since it is outside this range.

This leaves us with $x_2 = 1$, which means:

$$\cos^2 \theta = 1$$

Step 4: Solve for $\theta$

If $\cos^2 \theta = 1$, then $\cos \theta = \pm 1$.

We now solve for $\theta$ in the interval $0 \leq \theta \leq 2\pi$:

• $\cos \theta = 1$ occurs at $\theta = 0$ and $\theta = 2\pi$
• $\cos \theta = -1$ occurs at $\theta = \pi$

Step 5: Conclusion

The three solutions for $\theta$ are $0$, $\pi$, and $2\pi$.

Thus, there are 3 real solutions to the equation in the given interval.