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(C) Given the curve $y = 2 \cos x$.At $x = \frac{\pi}{4}$,the value of $y$ is $y = 2 \cos \left( \frac{\pi}{4} \right) = 2 \left( \frac{1}{\sqrt{2}} \

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$y - \sqrt{2} = -\sqrt{2} \left( x - \frac{\pi}{4} \right)$

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(C) Given the curve $y = 2 \cos x$.At $x = \frac{\pi}{4}$,the value of $y$ is $y = 2 \cos \left( \frac{\pi}{4} \right) = 2 \left( \frac{1}{\sqrt{2}} \right) = \sqrt{2}$.So,the point of tangency is $\left( \frac{\pi}{4}, \sqrt{2} \right)$.Now,find the derivative $\frac{dy}{dx} = \frac{d}{dx} (2 \cos x) = -2 \sin x$.The slope of the tangent at $x = \frac{\pi}{4}$ is $m = \left( \frac{dy}{dx} \right)_{x = \pi/4} = -2 \sin \left( \frac{\pi}{4} \right) = -2 \left( \frac{1}{\sqrt{2}} \right) = -\sqrt{2}$.The equation of the tangent line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.Substituting the values,we get $y - \sqrt{2} = -\sqrt{2} \left( x - \frac{\pi}{4} \right)$.

What is the equation of the tangent to the curve $y = 2 \cos x$ at $x = \pi / 4$?

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