The number of integral values of k for which | vedclass

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(B) The equation is of the form $a \cos x + b \sin x = c$,which has a solution if and only if $-\sqrt{a^2 + b^2} \leq c \leq \sqrt{a^2 + b^2}$.Here,$a

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$8$

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(B) The equation is of the form $a \cos x + b \sin x = c$,which has a solution if and only if $-\sqrt{a^2 + b^2} \leq c \leq \sqrt{a^2 + b^2}$.Here,$a = 7$,$b = 5$,and $c = 2k + 1$.Thus,$-\sqrt{7^2 + 5^2} \leq 2k + 1 \leq \sqrt{7^2 + 5^2}$.$-\sqrt{49 + 25} \leq 2k + 1 \leq \sqrt{49 + 25}$.$-\sqrt{74} \leq 2k + 1 \leq \sqrt{74}$.Since $\sqrt{74} \approx 8.602$,we have $-8.602 \leq 2k + 1 \leq 8.602$.Subtracting $1$ from all sides: $-9.602 \leq 2k \leq 7.602$.Dividing by $2$: $-4.801 \leq k \leq 3.801$.The integral values of $k$ are $\{-4, -3, -2, -1, 0, 1, 2, 3\}$.The total number of such integral values is $8$.

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

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