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Trigonometrical Equations
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જો $|k|\, = 5$ અને ${0^o} \le \theta \le {360^o}$, તો સમીકરણ $3\cos \theta + 4\sin \theta = k$ ની કેટલા ભિન્ન ઉકેલ શક્ય છે ?
A
$0$
B
$2$
C
$1$
D
અનંત
Solution
(b) $3\cos \theta + 4\sin \theta = 5\,\left[ {\frac{3}{5}\cos \theta + \frac{4}{5}\sin \theta } \right] = 5\cos (\theta – \alpha )$
where $\cos \alpha = \frac{3}{5}$, $\sin \alpha = \frac{4}{5}$
Now $3\cos \theta + 4\sin \theta = k$
$\therefore$ $5\cos (\theta – \alpha ) = k$
$\Rightarrow \cos (\theta – \alpha ) = \pm 1$
$ \Rightarrow $ $\theta – \alpha = {0^o},\,{180^o} $
$\Rightarrow \theta = \alpha ,\,{\rm{ }}{180^o} + \alpha $.
Standard 11
Mathematics
