Solve, for the equation 3 cos^2x + 10 sinx cosx = 0
when x is between 0 and 360 degrees ( inclusive of 0 and 360 degrees)
Pls show your workings...and explain too..tks a lot
2007-10-13 12:47:05 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
but the ans is 60, 270, 168.3 and 248.3....
2007-10-13 13:10:54 · update #1
it should be 60, 270, 168.3 and 348.3....
2007-10-13 13:11:22 · update #2
3 cos² x + 10 sinx cosx = 0
Factorize the shared (cos x):
(cos x)(3 cos x + 10 sinx) = 0
Hence either
(cos x) = 0
=> x=90°, 270°
OR:
(3 cos x + 10 sinx) = 0
=> 3 cos x = -10 sinx
=> -0.3 = tan x
=> x = 163.3° or 343.3°
x = 90°, 163.3°, 270°, 343.3° [SOLUTION]
2007-10-13 12:55:48 · answer #1 · answered by smci 7 · 0⤊ 0⤋
3cos^2x + 10sinxcosx = 0
cos x (3cosx + 10sinx) = 0
cos x = 0 x = 90 or 270
3cosx + 10sinx = 0
sinx /cosx = -3/10
tan x = -0.3 x = 180 - 16.7 = 163.3 (second quad)
or 360 - 16.7 = 343.3 (4th quad)
2007-10-13 12:59:03 · answer #2 · answered by norman 7 · 0⤊ 0⤋
pull out a cosine.
2007-10-13 12:57:49 · answer #3 · answered by ballin' no lie 2 · 0⤊ 0⤋