sin 3x + sin^2(x) = 2

Question

Solve the equation sin 3x + sin^2(x) = 2  for 0<x<360

Answer 1

sin(3x)+sin^2(x)=2

sin(2x+x)+ sin^2(x)

sin2xcosx+cos2xsinx+sin^2x

2sinxcos^2x+sinx(1-2sin^2x)+sin^2x

2sinx(1-sin^2x)+sinx-2sin^3x+sin^2x

2sinx-2sin^3x+sinx-2sin^3x+sin^2x

4sin^3x-3sinx-sin^2x+2=0

 

Say. sinx = y

 

4y^3-y^2-3y+2=0

Solve the polynomial equation

We get one root as y=-1

For other roots

Divide the polynomial by y+1

We get. 4y^2-5y+2

Once we find the roots of this polynomial

We get imaginary roots.

 

So, the only real solution is y=-1

So,  sinx=-1

x= 3*pi/2

 

so, no guessing of solutions here.

 

 

Answer 2

The idea here is to find an x-value that makes each term equal 1, and therefore, the sum equal to 2.

 

The sine function attains the values 1 and -1 at all n*pi/2 values, where n is an integer.  Since we require x to be between 0 and 2*pi (or as you have written it, 0 and 360 degrees), let's start with the guess of x₁= pi/2.

 

sin(3*x₁) + sin²(x₁) = sin(3*pi/2) + sin²(pi/2) = -1 + 1² = 0.  So the value x₁=pi/2 fails.

 

Next, trying a guess of x₂ = 2*pi/2 = pi would simply yield 0.

 

Next, try a guess of x₃ = 3*pi/2.

 

sin(3*x₃) + sin²(x₃) = sin(9*pi/2) + sin²(3*pi/2) = 1 + (-1)² = 2, as desired.

 

So, a solution to the equation is x=3*pi/2.