Solve the solutions in the interval [0,2p)

Question

a) 2sin x = -3

b) -5 tan x = 5

c)√2 sec x - 2 = 0

d) sin 2x + √2 sin x = 0

e) cos2x = - cos x

Answer 1

Solutions in [0,2π) for each:

 

a) 2 sin x = -3

 

sin x = -3/2

 

Since -3/2 ∉ [-1,1], it is not the sine of any angle so there are no solutions.

 

b) -5 tan x = 5

 

tan x = -1

 

x = nπ + tan^-1 -1 = (n-1/4)π , n ∈ Z.

 

So the solution set is {3π/4, 7π/4}.

 

c) √2 sec x - 2 = 0

 

√2 sec x = 2

 

sec x = √2

 

cos x = 1/√2 = √2 / 2

 

x = 2nπ ± cos^-1 (√2 / 2)  = (2n±1/4)π, n ∈ Z.

 

So the solution set is {π/4, 7π/4}.

 

d) sin 2x + √2 sin x = 0

 

2 sin x cos x  + √2 sin x = 0

 

(2 cos x + √2) sin x = 0

 

cos x = -√2 / 2 or sin x = 0

 

x = 2nπ ± cos^-1 (-√2 / 2) or x = 2nπ ± sin^-1 0, n ∈ Z.

 

x = 2nπ ± 3π/4 = (2n±3/4)π or x = 2nπ, n ∈ Z.

 

So the solution set is {0, 3π/4, π, 5π/4}

 

e) cos 2x = -cos x

 

2 cos² x + cos x = 0

 

(2 cos x + 1)cos x = 0

 

cos x = -1/2 or cos x = 0

 

x = 2nπ ± cos^-1 -1/2 or x = 2nπ ± cos^-1 0

 

x = 2nπ ± 2π/3 or x = 2nπ ± π/2, n ∈ Z.

 

x = (2n ± 2/3)π or x = (2n ± 1/2)π, n ∈ Z.

 

So the solution set is {π/2, 2π/3, 3π/2, 4π/3}