Find the solutions for this

Question

0º < x < 360º sin 2x - cos x = 0?

Answer 1

Given equation : sin 2x - cosx = 0

Interval is 0 < x < 360

Substitute : sin 2x = 2 sinx cosx

2 sinx cosx - cosx = 0

( 2 sinx - 1 ) cosx  = 0

By using zero product property : If AB = 0 then A = 0 , B = 0

2 sinx - 1 = 0   and    cosx  = 0

Solve above equations separately.

Part1 : 2 sinx - 1 = 0

2 sinx = 1

sinx = 1/2

sinx = sin(30^o)

α = 30^o

General solution: x = 180^on +(-1)ⁿα, where n is an integer.

x = 180^on +(-1)ⁿ30^o

If n = 0

x = 180^o*0 +(-1)⁰30^o

x = 30^o   ---->  This angle is in given interval 0 < x < 360.
So x = 30^o  is one solution.

If n = 1

x = 180^o*1 +(-1)¹30^o

x = 180^o - 30^o

x = 150^o   ---->  This angle is in given interval 0 < x < 360.

So x = 150^o  is one solution.

If n = 2

x = 180^o*2 +(-1)²30^o

x = 360^o + 30^o

x = 390^o   ---->  This angle is out of range in given interval 0 < x < 360

As n increases x also increases .So for n > 2 we will get no solutions.

sinx = 1/2 ⇒ x = {  30^o , 150^o }

Part 2 : cosx  = 0

cosx  = cos 90

General solution: x = 360^on ± 90, where n is an integer.

If n = 0

x = 360^on ± 90

x = 360^o*0 ± 90 = ± 90

x =  90^o   ---->  This angle is in given interval 0 < x < 360.

So x = 90^o  is one solution.

x =  - 90^o   ---->  This angle is out of range in given interval 0 < x < 360

If n = 1

x = 360^o*1 ± 90 = 450 , 270

x =  270^o   ---->  This angle is in given interval 0 < x < 360.

So x = 270^o  is one solution.

x =  450^o   ---->  This angle is out of range in given interval 0 < x < 360

As n increases x also increases .So for n > 1 we will get no solutions.

cosx  = 0 ⇒ x = { 90^o , 270^o }

Final solution is

x = { 30^o , 90^o , 150^o , 270^o }