Question

Solve the following equations:
sin 2θ – cos 2θ – sin θ + cos θ = θ

Answer 1

sin 2θ – cos 2θ – sin θ + cos θ = θ

`2cos ((2theta + theta)/2) sin ((2theta - theta)/2) - 2 sin ((2theta + theta)/2) sin ((theta - 2theta)/2)` = 0

`2cos ((3theta)/2) * sin  (theta/2) - 2sin ((3theta)/2)  sin  (- theta/2)` = 0

`2cos ((3theta)/2) * sin  (theta/2) + 2sin ((3theta)/2)  sin  (theta/2)` = 0

`2sin  theta/2 [cos ((3theta)/2) + sin  ((3theta)/2)]` = 0

`2 sin  theta/2` = 0 or `cos  ((3theta)/2) + sin  ((3theta)/2)` = 0

`sin  theta/2` = 0 or `cos  ((3theta)/2) = - sin  ((3theta)/2)`

`sin  theta/2` = 0 or `(sin  ((3theta)/2))/(cos  ((3theta)/2))` = – 1

`sin  theta/2` = 0 or `tan  ((3theta)/2)` = – 1

To find the general solution of `sin  theta/2` = 0

The general solution is

`theta/2` = nπ, n ∈ Z

 θ = 2nπ, n ∈ Z
To find the general solution of `tan  ((3theta)/2)` = – 1

`tan  ((3theta)/2)` = – 1

`tan  ((3theta)/2) = tan (pi - pi/4)`

`tan  ((3theta)/2) = tan ((4pi - pi)/4)`

`tan  ((3theta)/2) = tan  ((3pi)/4)`

The general solution is

`(3theta)/2 = "n" + pi/4`, n ∈ Z

θ = `(2"n"pi)/3 + (2pi)/(3 xx 4)`, n ∈ Z

θ = `(2"n"pi)/3 + pi/6`, n ∈ Z

∴ The required solutions are

θ = 2nπ, n ∈ Z

or

θ = `(2"n"pi)/3 + pi/6`, n ∈ Z