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First, I put $t = \sin x - \cos x$,

eq1 = (3 - Cos[4x]) ( Sin[x] - Cos[x]) - 2 == 0; eq2 = t == Sin[x] - Cos[x]; Eliminate[ TrigExpand[ {eq1, eq2}], x] 

I receive

2 t - 2 t^3 + t^5 == 1

And then, I solve

Solve[ 2 t - 2 t^3 + t^5 == 1, Reals] 

finally

Reduce[ -Cos[x] + Sin[x] == 1, x, Reals] (C[1] ∈ Integers && x == π/2 + 2 π C[1]) || (C[1] ∈ Integers && x == π + 2 π C[1]) 

This is my solution by hand, I put at http://math.stackexchange.com/questions/218381/how-to-solve-this-trigonometric-equation/218496#218496

First, I put $t = \sin x - \cos x$,

eq1 = (3 - Cos[4x]) ( Sin[x] - Cos[x]) - 2 == 0; eq2 = t == Sin[x] - Cos[x]; Eliminate[ TrigExpand[ {eq1, eq2}], x] 

I receive

2 t - 2 t^3 + t^5 == 1

And then, I solve

Solve[ 2 t - 2 t^3 + t^5 == 1, Reals] 

finally

Reduce[ -Cos[x] + Sin[x] == 1, x, Reals] (C[1] ∈ Integers && x == π/2 + 2 π C[1]) || (C[1] ∈ Integers && x == π + 2 π C[1]) 

This is my solution by hand, I put at https://math.stackexchange.com/questions/218381/how-to-solve-this-trigonometric-equation/218496#218496

First, I put $t = \sin x - \cos x$,

eq1 = (3 - Cos[4*x])*(Sin[x] - Cos[x]) - 2 == 0; eq2 = t == Sin[x] - Cos[x] Eliminate[TrigExpand[{eq1, eq2}], x] 

I receive

2 t - 2 t^3 + t^5 == 1

And then, I solve

Solve[2 t - 2 t^3 + t^5 == 1, Reals] 

finally

Reduce[-Cos[x] + Sin[x] == 1, x, Reals] (C[1] \[Element] Integers &&  x == \[Pi]/2 + 2 \[Pi] C[1]) || (C[1] \[Element] Integers &&  x == \[Pi] + 2 \[Pi] C[1]) 

This is my solution by hand, I put at http://math.stackexchange.com/questions/218381/how-to-solve-this-trigonometric-equation/218496#218496

First, I put $t = \sin x - \cos x$,

eq1 = (3 - Cos[4x]) ( Sin[x] - Cos[x]) - 2 == 0; eq2 = t == Sin[x] - Cos[x]; Eliminate[ TrigExpand[ {eq1, eq2}], x] 

I receive

2 t - 2 t^3 + t^5 == 1

And then, I solve

Solve[ 2 t - 2 t^3 + t^5 == 1, Reals] 

finally

Reduce[ -Cos[x] + Sin[x] == 1, x, Reals] (C[1] ∈ Integers && x == π/2 + 2 π C[1]) || (C[1] ∈ Integers && x == π + 2 π C[1]) 

This is my solution by hand, I put at http://math.stackexchange.com/questions/218381/how-to-solve-this-trigonometric-equation/218496#218496

First, I put $t = \sin x - \cos x$,

eq1 = (3 - Cos[4*x])*(Sin[x] - Cos[x]) - 2 == 0; eq2 = t == Sin[x] - Cos[x] Eliminate[TrigExpand[{eq1, eq2}], x] 

I receive

2 t - 2 t^3 + t^5 == 1

And then, I solve

Solve[2 t - 2 t^3 + t^5 == 1, Reals] 

finally

Reduce[-Cos[x] + Sin[x] == 1, x, Reals] (C[1] \[Element] Integers && x == \[Pi]/2 + 2 \[Pi] C[1]) || (C[1] \[Element] Integers && x == \[Pi] + 2 \[Pi] C[1]) 

First, I put $t = \sin x - \cos x$,

eq1 = (3 - Cos[4*x])*(Sin[x] - Cos[x]) - 2 == 0; eq2 = t == Sin[x] - Cos[x] Eliminate[TrigExpand[{eq1, eq2}], x] 

I receive

2 t - 2 t^3 + t^5 == 1

And then, I solve

Solve[2 t - 2 t^3 + t^5 == 1, Reals] 

finally

Reduce[-Cos[x] + Sin[x] == 1, x, Reals] (C[1] \[Element] Integers && x == \[Pi]/2 + 2 \[Pi] C[1]) || (C[1] \[Element] Integers && x == \[Pi] + 2 \[Pi] C[1]) 

This is my solution by hand, I put at http://math.stackexchange.com/questions/218381/how-to-solve-this-trigonometric-equation/218496#218496