有兩條三角方程唔識解，好急呀，唔該曬！！

1)3cos²x + sinxcosx - 1 = 0
3cos²x + sinxcosx - (sin²x + cos²x) = 0
2cos²x + sinxcosx - sin²x = 0
(2cosx - sinx) (cosx + sinx) = 02cosx - sinx = 0
tanx = 2
x = nπ + arctan2 = nπ + 1.10715
Sub. n = 0 , 1 :
x = 1.10715 or 4.24874 for 0 ≤ x < 2π.orcosx + sinx = 0
tanx = - 1
x = nπ + arctan -1 = nπ - π/4
Sub. n = 1 , 2 :
x = 3π/4 or 7π/4 for 0 ≤ x < 2π.

2)2tanx + 4 = 3√2 secx
2 sinx / cosx + 4 = 3√2 / cosx
2sinx + 4cosx = 3√2
√(2² + 4²) sin (x + φ) = 3√2(were sinφ = 4/√(2² + 4²) , cosφ = 2/√(2² + 4²)
i.e. tanφ = 2 ,
φ = arctan2 for sinφ is +ve and cosφ is +ve and 0 ≤ φ < 2π ) sin (x + φ) = 3√2 / (2√5)
x + φ = nπ + (-1)ⁿ arcsin√0.9
x = nπ + (-1)ⁿ arcsin√0.9 - arctan2
Sub. n = 0 , 1 :
x = 0.1419 or π/4 for 0 ≤ x < 2π.


2012-10-12 00:37:28 補充：
Alternatively :

2tanx + 4 = 3√2 secx
2 sinx / cosx + 4 = 3√2 / cosx
2sinx + 4cosx = 3√2
2sinx ± 4√(1 - sin²x) = 3√2

2012-10-12 00:37:33 補充：
16(1 - sin²x) = ( 3√2 - 2sinx)²
16 - 16sin²x = 18 - 12√2 sinx + 4sin²x
10sin²x - 6√2 sinx + 1 = 0
sinx = (3√2 ± 2√2) / 10

sinx = √2 / 2
or √2 / 10

x = π/4 , 3π/4 (rejected since R.H.S. > 0 but L.H.S. < 0 )
or x = 0.1419 , 2.9997 (rejected since R.H.S. > 0 but L.H.S. < 0 )