AP EAMCET · Maths · Trigonometric Equations
General solution of \(4 \sin ^2(x)-4 \sin (x)+1=0\) is
- A \(x=2 n \pi \pm \frac{\pi}{3}, n \in Z\)
- B \(x=n \pi+(-1)^n \frac{\pi}{3}, n \in Z\)
- C \(x=n \pi+(-1)^n \frac{\pi}{6}, n \in Z\)
- D \(x=n \pi+(-1)^n \frac{\pi}{4}, n \in Z\)
Answer & Solution
Correct Answer
(C) \(x=n \pi+(-1)^n \frac{\pi}{6}, n \in Z\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} 4 \operatorname{Sin}^2 x-4 \operatorname{Sin} x+1 & =0 \\ (2 \operatorname{Sin} x-1)^2 & =0 \\ 2 \operatorname{Sin} x-1 & =0 \\ \operatorname{Sin} x & =\frac{1}{2} \\ x & =n \pi+(-1)^n \cdot \frac{\pi}{6}, n \in \mathbf{Z} \end{aligned}\) Hence, option (c) is…
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