What is k if f(x) = f'(x) for f(x) = sin x + k*cos x

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The function f(x) = sin x + k*cos x. The derivative of f(x) is f'(x) = cos x - k*sin x

As f(x) = f'(x)

=> sin x + k*cos x = cos x - k*sin x

=> k*(sin x + cos x) = cos x - sin x

=> k = (cos x - sin x)/(sin x + cos x)

There are an infinite number of values that k can take, for example for x = 30 degrees, k = `2 - sqrt 3` , for x = 45 degrees k = 0, and for x = 90 degrees k = -1.

**There are an infinite number of values of x for which f(x) can be equal to f'(x) for the function f(x) = sin x + k*cos x**

The function f(x) = sin x + k cos x. The derivative of f(x) is

f'(x) = cos x - ksin x

given f(x) = f'(x)

sin x + k cos x = cos x - k sin x

k (sin x + cos x) = cos x - sin x

k = (cos x - sin x)/(sin x + cos x)

k=(1-tanx)/(1+tax)

k=tan((pi/4)-x)

since it is transcedental /periodic equation .It has infinite no of real solution provided `x!=-((4n+1)pi)/4`

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