Checking whether the signal is energy or power
∫−∞ ∞ ∣x(t)∣2∫−∞ ∞ (sin(2t)+3cos(4t))2dt=5t−18sin(4t)−12cos(6t)+32cos(2t)+916sin(8t)+C=∞\int _{-\infty \:}^{\infty \:}|x(t)|^2\\ \int _{-\infty \:}^{\infty \:}\left(\sin \left(2t\right)+3\cos \left(4t\right)\right)^2dt\\ =5t-\frac{1}{8}\sin \left(4t\right)-\frac{1}{2}\cos \left(6t\right)+\frac{3}{2}\cos \left(2t\right)+\frac{9}{16}\sin \left(8t\right)+C\\ =\infin∫−∞∞∣x(t)∣2∫−∞∞(sin(2t)+3cos(4t))2dt=5t−81sin(4t)−21cos(6t)+23cos(2t)+169sin(8t)+C=∞
The energy of the signal is infinite hence the signal is power.
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