Trigonometric Equations TG EAPCET Previous Year Chapters wise Questions

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To solve this problem, let's consider the solution sets of the given equations step by step.Set A:We start with the equation: $\cos^2 x = \cos^2 \frac{\pi}{6}$.Since $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$, we have $\cos^2 \frac{\pi}{6} = \frac{3}{4}$.The solutions to this equation are:$x = n\pi \pm \frac{\pi}{6},$ where $n \in \mathbb{Z}$Set B:We consider the equation: $\cos^2 x = \log_{16} P$.Given $P + \frac{16}{P} = 10$, finding $P$ involves solving the quadratic equation $p^2 - 10p + 16 = 0$.Factoring gives: $(p-8)(p-2)=0$, so $p=8$ or $p=2$.Thus, $\cos^2 x = \log_{16} 8$ or $\cos^2 x = \log_{16} 2$.For $p=8, \log_{16} 8 = \frac{3}{4} (.$ since $\frac{3}{4} = \cos^2 \left(\frac{\pi}{6}\right) )$.For $p=2, \log_{16} 2 = \frac{1}{4}$.Hence, $\cos^2 x = \frac{3}{4}$ aligns with $\cos^2 x = \cos^2 \frac{\pi}{6}$.For $\cos^2 x = \frac{1}{4}$, the solutions are: $x = 2n\pi \pm \frac{\pi}{3}$ and $x = 2n\pi \pm \frac{2\pi}{3},$ where $n \in \mathbb{Z}$Finding $B-A$ :Set $B$ includes solutions for both $\cos^2 x = \frac{1}{4}$ and $\cos^2 x = \frac{3}{4}$.Set $A$ only includes solutions for $\cos^2 x = \frac{3}{4}$.Therefore, the difference $B-A$ consists of solutions only when $\cos^2 x = \frac{1}{4}$, which are:$B-A = \{x \in \mathbb{R} \mid x=2n\pi \pm \frac{\pi}{3}, 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z}\}$