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General solution of 2sinx.sin2x = 1

User Doxylee
by
8.1k points

1 Answer

2 votes

Givens

sin(2x) = 2 * sin(x) * cos(x)

Equation

sin(x)*sin2(X) = 2sin(x) * 2sin(x)*cos(x) = 1

4*sin^2(x)* cos(x) = 1

sin^2(x) = 1 - cos^2(x)

4(1 - cos^2(x) ) * cos(x) = 1 Divide by 4

(1 - cos(x) ) *cos(x) = 1/4

cos(x) - cos^3(x) = 1/4

Let y = cos(x)

y - y^3 = 1/4

This is a cubic. I have to use a calculator's built in formula to solve it.

a = -1

b = 0

c = 1

d = - 1/4

Solution

y1 = - 1.107. Extraneous cos(x) can't be less than 1.

y2 = 0.8376

y3 = 0.2696

but cos(x) = y

cos-1(0.8376) = 33.11 degrees.

cos-1(0.2696) = 74.36 degrees.

The cosine is also plus in quad 4.

x = 360 - 33.11 = 326.89

x = 360 - 74.36 = 285.64

The graph of the cubic is below. It pretty much confirms what I found.

General solution of 2sinx.sin2x = 1-example-1
User Erikvm
by
8.3k points

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