Question

If general solution of $\cos^2\theta - 2\sin\theta + \frac{1}{4} = 0$ is $\theta = \frac{n\pi}{A} + (-1)^n\frac{\pi}{B}, n \in Z$ then $A+B$ has the value

• A. 7
• B. 6
• C. 1
• D. -7

Answer 1

Answer: (A)

7

Answer 2

1. Rewrite the equation using the Pythagorean identity:

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

   Substituting into the equation:

   $$1 - \sin^2\theta - 2\sin\theta + \frac{1}{4} = 0 \quad \Rightarrow \quad -\sin^2\theta - 2\sin\theta + \frac{5}{4} = 0$$

   Multiply by -1:

   $$\sin^2\theta + 2\sin\theta - \frac{5}{4} = 0$$

2. Solve the quadratic equation in $\sin\theta$:

   $$\sin\theta = \frac{-2 \pm \sqrt{4 + 5}}{2} = \frac{-2 \pm 3}{2}$$

   The valid solution is:

   $$\sin\theta = \frac{1}{2}.$$

3. The general solutions for $\sin\theta = \frac{1}{2}$ are:

   $$\theta = \frac{\pi}{6} + 2\pi k \quad \text{and} \quad \theta = \frac{5\pi}{6} + 2\pi k \quad (k \in \mathbb{Z})$$

4. These can be written in the form:

   $$\theta = \pi k + (-1)^k \frac{\pi}{6} \quad (k \in \mathbb{Z})$$

   Comparing with the given form:

   $$\theta = \frac{n\pi}{A} + (-1)^n\frac{\pi}{B},$$

   we deduce $A = 1$ and $B = 6$.

5. Therefore,

   $$A + B = 1 + 6 = 7.$$

Core Explanation:
Rewrite using $1-\sin^2\theta$, solve for $\sin\theta=1/2$, express general solutions as $\theta = \pi k + (-1)^k\pi/6$ which corresponds to $A=1, B=6$. So, $A+B=7$.