If |k| = 5 and 0^o le theta le 360^o,then the | vedclass

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(B) The given equation is $3\cos 	heta + 4\sin 	heta = k$.We can write this as $5 \left( \frac{3}{5}\cos 	heta + \frac{4}{5}\sin 	heta ight) = k

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(B) The given equation is $3\cos 	heta + 4\sin 	heta = k$.We can write this as $5 \left( \frac{3}{5}\cos 	heta + \frac{4}{5}\sin 	heta ight) = k$.Let $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$. Then the equation becomes $5\cos(	heta - \alpha) = k$.Given $|k| = 5$,we have $k = 5$ or $k = -5$.Case $1$: If $k = 5$,then $\cos(	heta - \alpha) = 1$. In the interval $0^o \le 	heta \le 360^o$,there is exactly one solution for $	heta - \alpha = 0^o$,which is $	heta = \alpha$.Case $2$: If $k = -5$,then $\cos(	heta - \alpha) = -1$. In the interval $0^o \le 	heta \le 360^o$,there is exactly one solution for $	heta - \alpha = 180^o$,which is $	heta = 180^o + \alpha$.In both cases,for a fixed $k$ such that $|k|=5$,there is exactly one solution for $	heta$ in the given range. However,the question asks for the number of solutions for the equation $3\cos 	heta + 4\sin 	heta = k$ where $|k|=5$. Since $k$ can be $5$ or $-5$,and for each value there is one solution,the total number of solutions is $2$.

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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