The solution to 4sin2xcosx+sin3x=sin<3,14+x> is x = nπ, where n = ±{0, 0.5, 1, 1.5....}
How to solve the equation?
The equation is given as:
4sin2xcosx+sin3x=sin<3,14+x>
Rewrite properly as:
4sin(2x)cos(x)+sin(3x)=sin(3.14+x)
Next, we split the equation as follows:
y = 4sin(2x)cos(x)+sin(3x)
y = sin(3.14+x)
Next, we plot the graph of both equations (see attachment)
From the attached graph, we have:
x = nπ, where n = ±{0, 0.5, 1, 1.5....}
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