What is expand 6(-e^{3x}(2sin(2x)-3cos(2x))) ?

The solution to expand 6(-e^{3x}(2sin(2x)-3cos(2x))) is -12e^{3x}sin(2x)+18e^{3x}cos(2x)

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expand 6(-e^{3x}(2sin(2x)-3cos(2x)))

expand

6 (- e^{3 x} (2 sin (2 x) - 3 cos (2 x)))

- 12 e^{3 x} sin (2 x) + 18 e^{3 x} cos (2 x)

Solution steps

6 (- e^{3 x} (2 sin (2 x) - 3 cos (2 x)))

Remove parentheses:

(- a) = - a

= - 6 e^{3 x} (2 sin (2 x) - 3 cos (2 x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 6 e^{3 x}, b = 2 sin (2 x), c = 3 cos (2 x)

= - 6 e^{3 x} \cdot 2 sin (2 x) - (- 6 e^{3 x}) \cdot 3 cos (2 x)

Apply minus-plus rules

- (- a) = a

= - 6 \cdot 2 e^{3 x} sin (2 x) + 6 \cdot 3 e^{3 x} cos (2 x)

Simplify

- 6 \cdot 2 e^{3 x} sin (2 x) + 6 \cdot 3 e^{3 x} cos (2 x) : - 12 e^{3 x} sin (2 x) + 18 e^{3 x} cos (2 x)

= - 12 e^{3 x} sin (2 x) + 18 e^{3 x} cos (2 x)

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c H (r (o, s, x), o, s, x) a d r (o, s, x) a d os (x - 2 x = 6)

\frac{12.9}{( s + 1.1 ) ( s + 2 )}

6 sin (x) (x + 1)

What is expand 6(-e^{3x}(2sin(2x)-3cos(2x))) ?
The solution to expand 6(-e^{3x}(2sin(2x)-3cos(2x))) is -12e^{3x}sin(2x)+18e^{3x}cos(2x)