Trigonometric Equations Question 39
Question: If $ |k|,=5 $ and $ 0^{o}\le \theta \le 360^{o} $ , then the number of different solutions of 3 $ \cos \theta +4\sin \theta =k $ is
Options:
A) Zero
B) Two
C) One
D) Infinite
Show Answer
Answer:
Correct Answer: B
Solution:
- $ 3\cos \theta +4\sin \theta =5,[ \frac{3}{5}\cos \theta +\frac{4}{5}\sin \theta ]=5\cos (\theta -\alpha ) $ where $ \cos \alpha =\frac{3}{5} $ , $ \sin \alpha =\frac{4}{5} $ Now $ 3\cos \theta +4\sin \theta =k $ \ $ 5\cos (\theta -\alpha )=k\Rightarrow \cos (\theta -\alpha )=\pm 1 $
$ \Rightarrow $ $ \theta -\alpha =0^{o},,180^{o}\Rightarrow \theta =\alpha ,,180^{o}+\alpha $ .
BETA
