Trigo questions

je261 SG

1) 1+ 2cos(3x/2 +75deg) = 0

Ans: 30, 110, 270, 350

I only managed to get the first 2, how do i get 270 & 350?

2) sec(x/2 - 107deg) = 2

Ans: 94, 334

I only managed to get 334, how do i get 94?

Thanks in advance!

HyuugaNeji SG

I give you the range.

If they want x to be between 0 and 360,

then 0<x<360

0<3x/2<540
75<3x/2 +75<615

Happy?

dreamer87 SG

1+ 2cos(3x/2 +75deg) = 0
cos(3x/2 + 75) = -1/2
cos(alpha) = -1/2, 0 < alpha < 90
alpha = 60 deg

cos(alpha) = -1/2 ==> cos is negative ==> 2nd and 3rd quarter
3x/2 + 75 = 180 - alpha, 180+alpha, 360 + 180 + alpha, 360 + 180 - alpha
3x/2 + 75 = 120, 240, 600, 480
3x/2 = 45, 165, 525, 405
x = 30, 110, 350, 270

sec(x/2 - 107deg) = 2
1/cos(x/2 - 107deg) = 2
cos(x/2 - 107deg) = 1/2
cos(alpha) = 1/2
alpha = 60

cos(alpha) = 1/2 ==> first and fourth quarter
0<x<360 ==> 0<x/2<180 ==> -107 < x/2-107<73

Therefore,
x/2 - 107deg = -alpha ,alpha,
x/2 - 107deg = -60, 60
x/2 = 47,167
x = 94, 334

Hope it helps!

Darkness_hacker99 SG

thx for sharing

je261 SG

Thanks dreamer87, your detailed working has helped alot!