If |k|, = 5 and {0^o} le theta le {360^o}, th | vedclass

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(b) $3\cos 	heta + 4\sin 	heta = 5\,\left[ {\frac{3}{5}\cos 	heta + \frac{4}{5}\sin 	heta } ight] = 5\cos (	heta - \alpha )$ 

where $\c

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Two

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(b) $3\cos 	heta + 4\sin 	heta = 5\,\left[ {\frac{3}{5}\cos 	heta + \frac{4}{5}\sin 	heta } ight] = 5\cos (	heta - \alpha )$ 

where $\cos \alpha = \frac{3}{5}$, $\sin \alpha = \frac{4}{5}$ 

Now $3\cos 	heta + 4\sin 	heta = k$  

$	herefore$ $5\cos (	heta - \alpha ) = k$

$\Rightarrow \cos (	heta - \alpha ) = \pm 1$

$ \Rightarrow $ $	heta - \alpha = {0^o},\,{180^o} $

$\Rightarrow 	heta = \alpha ,\,{m{ }}{180^o} + \alpha $.

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

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