rachelmaddie
Full Member
- Joined
- Aug 30, 2019
- Messages
- 851
I need this checked please.
Rewrite the equation;
tan2x (sec2x - 1) + 2sec2x - 2 =0
Lets name: sin(x) = s, cos(x)=c, then
tan(x) = s/c, sec(x) = 1/c, and s2 + c2 =1.
The equation in terms of new variables s and c:
s2/c2 (1/c2 -1) + 2/c2 -2 =0
s2/c2 (1 - c2)/c2 + 2/c2 -2 =0
s4/c4 + 2/c2 -2 =0
s4 + 2c2 -2c4 =0
Replace: s4 = (1- c2)2 =1 - 2c2 + c4
1 - 2c2 + c4 + 2c2 -2c4 =0
1 - c4 =0
c = 1 or c = -1
cos(x) = 1 or cos(x) = -1
Answer: x = 0, x= π
Rewrite the equation;
tan2x (sec2x - 1) + 2sec2x - 2 =0
Lets name: sin(x) = s, cos(x)=c, then
tan(x) = s/c, sec(x) = 1/c, and s2 + c2 =1.
The equation in terms of new variables s and c:
s2/c2 (1/c2 -1) + 2/c2 -2 =0
s2/c2 (1 - c2)/c2 + 2/c2 -2 =0
s4/c4 + 2/c2 -2 =0
s4 + 2c2 -2c4 =0
Replace: s4 = (1- c2)2 =1 - 2c2 + c4
1 - 2c2 + c4 + 2c2 -2c4 =0
1 - c4 =0
c = 1 or c = -1
cos(x) = 1 or cos(x) = -1
Answer: x = 0, x= π