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
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#
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