What is expand e^x(3cos(2x)+sin(2x)) ?

The solution to expand e^x(3cos(2x)+sin(2x)) is 3e^xcos(2x)+e^xsin(2x)

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expand e^x(3cos(2x)+sin(2x))

expand

e^{x} (3 cos (2 x) + sin (2 x))

3 e^{x} cos (2 x) + e^{x} sin (2 x)

Solution steps

e^{x} (3 cos (2 x) + sin (2 x))

Apply the distributive law:

a (b + c) = ab + a c

a = e^{x}, b = 3 cos (2 x), c = sin (2 x)

= e^{x} \cdot 3 cos (2 x) + e^{x} sin (2 x)

= 3 e^{x} cos (2 x) + e^{x} sin (2 x)

(z^{n} - e^{j θ}) (z^{n} - e^{- j θ})

\frac{3}{( 2 - z ) ^{2}}

(n - r)! (n + 1 - r)

\frac{3 x - 2}{x ( x - 2 ) ( x + 1 ) ( x + 3 )}

\frac{- x + 1}{5 ( x + 2 )}

What is expand e^x(3cos(2x)+sin(2x)) ?
The solution to expand e^x(3cos(2x)+sin(2x)) is 3e^xcos(2x)+e^xsin(2x)