Hi, looking for another piece of the puzzle,
Find all the angles between 0 deg and 180 deg which satisfy
cot x - 2 cos 2x = 2
My working, (cos x / sin x) - 2 (2 cos^2 x -1) - 2 = 0
(cos x / sin x) - 4 cos ^2 x = 0
Multiply by tan x => 1 - 4 sin x cos x = 0
2 sin 2x = 1
sin 2x = 0.5 , basic angle is 30 deg.
x = (15, 75) deg
There is another 90 deg in the answer and I am looking for it. Is my working correct? Thanks. jt
Originally posted by Jt061952:Hi, looking for another piece of the puzzle,
Find all the angles between 0 deg and 180 deg which satisfy
cot x - 2 cos 2x = 2
My working, (cos x / sin x) - 2 (2 cos^2 x -1) - 2 = 0
(cos x / sin x) - 4 cos ^2 x = 0
Multiply by tan x => 1 - 4 sin x cos x = 0
2 sin 2x = 1
sin 2x = 0.5 , basic angle is 30 deg.
x = (15, 75) deg
There is another 90 deg in the answer and I am looking for it. Is my working correct? Thanks. jt
(cos x / sin x) - 4 cos ^2 x = 0... x(sinx)
cos x - 4 cos^2 x sinx = 0
cos x ( 1- 4sinxcosx)=0
cos x =0 or sin2x = 0.5
You lost the cos x -=0 which lead to x =90
Thanks Mike, careless of me....
Originally posted by Jt061952:Thanks Mike, careless of me....
Try not to multiply or divide any of the variables ( except numbers ), and factorise them whenever you could instead.