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4sin2xcosx+sin3x=sin<3,14+x>

Sagot :

MrRoyal

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....}

Read more about trigonometry functions at:

https://brainly.com/question/8120556

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View image MrRoyal
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