Number of integral values of k , for which equation 7cos x + 5sin x = 2k + 1 has a solution, is

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Trigonometrical Equations

hard

The number of integral values of $k$, for which the equation $7\cos x + 5\sin x = 2k + 1$ has a solution, is

$4$

$8$

$10$

$12$

(IIT-2002)

Solution

(b) $ – \sqrt {{7^2} + {5^2}} \le (7\cos x + 5\sin x) \le \sqrt {{7^2} + {5^2}} $
So, for solution $ – \sqrt {74} \le (2k + 1) \le \sqrt {74} $
or $ – 8.6 \le (2k + 1) \le 8.6$ or $ – 9.6 \le 2k \le 7.6$
or $ – 4.8 \le k \le 3.8.$ So, integral values of k are $ – 4,\, – 3,\, – 2,\, – 1,\,0,\,1,\,2,\,3$ $(eight\, values).$

Standard 11

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