Answer to Question #163513 in Civil and Environmental Engineering for Bonevie tutor

Question

Simple trigonometry equation

Solve trigonometric equation

1) 3 tan² x - 1 = 0

2) 2 cos x + 1 = 0

3) 2cos²-cos x - 1 = 0

4) 2cos² x - 3cos x + 1 = 0

Answer 1

3 tan² x - 1 = 0

(√3 tan x + 1). (√3 tan x - 1) = 0

√3 tan x + 1 = 0

√3 tan x = -1

tan x = -1/√3

tan x = -1/3 √3

x = 150° tan 330°

√3 tan x - 1 = 0

√3 tan x = 1

tan x = 1/√3

tan x = 1/3 √3

x = 30° tan 210°

2 cos x + 1 = 0

2cos(x) +1 =0

2cos(x) + 1-1=0-1

2cos(x)= -1

2cos(x) = -1

  2          2

2cos²-cos x - 1 = 0

#2cos^2x-2cosx+cosx-1=0#

or #2cosx(cosx-1)+1(cosx-1)=0#

or #(2cosx+1)(cosx-1)=0#

interval #[0,2pi)# #x=(2pi)/3# or #(4pi)/3#

 #cosx-1=# i.e. #cosx=1# and in given interval #x=0#.

Hence solution is #{0,(2pi)/3,(4pi)/3}#

 2cos² x - 3cos x + 1 = 0

#2t^2 - 3t + 1 = 0# .

#2t^2 - 2t - t + 1= 0#

#2t(t - 1) - 1(t - 1) = 0#

#(2t - 1)(t - 1) = 0#

#t = 1/2 and 1#

#t# with #cosx#.

#cosx = 1/2 and cosx = 1#

#x = pi/3, (5pi)/3#