The equation 3 cos x + 4 sin x = 6 has | vedclass

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(D) The given equation is $3 \cos x + 4 \sin x = 6$. We know that the range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2

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(D) The given equation is $3 \cos x + 4 \sin x = 6$. We know that the range of the expression $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$. Here,$a = 3$ and $b = 4$. So,the range is $[-\sqrt{3^2 + 4^2}, \sqrt{3^2 + 4^2}] = [-\sqrt{9 + 16}, \sqrt{9 + 16}] = [-5, 5]$. Since the value $6$ lies outside the range $[-5, 5]$,the equation $3 \cos x + 4 \sin x = 6$ has no solution.

The equation $3 \cos x + 4 \sin x = 6$ has

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