SOLUTION: Find all the angles between 0 and 360 degrees, i) 4(cos)^2(theta) = 4(sin)theta + 1 ii) 2 sin.x + 3 cos.x = 0 *Please answer as soon as possible bro :)

Algebra ->  Trigonometry-basics -> SOLUTION: Find all the angles between 0 and 360 degrees, i) 4(cos)^2(theta) = 4(sin)theta + 1 ii) 2 sin.x + 3 cos.x = 0 *Please answer as soon as possible bro :)      Log On


   

Question 465567: Find all the angles between 0 and 360 degrees,
i) 4(cos)^2(theta) = 4(sin)theta + 1
ii) 2 sin.x + 3 cos.x = 0
*Please answer as soon as possible bro :)
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find all the angles between 0 and 360 degrees,
i) 4(cos)^2(theta) = 4(sin)theta + 1
ii) 2 sin.x + 3 cos.x = 0
...
i) 4(cos)^2(theta) = 4(sin)theta + 1
use x instead of theta
4cos^2x=4sinx+1
4(1-sin^2x)=4sinx+1
4-4sin^2x=4sinx+1
4sin^2x+4sinx-3=0
(2sinx-1)(2sinx+3)=0
..
2sinx-1=0
sinx=1/2
x=30º and 135º (quadrants I and II) (reference angle=30º
..
2sinx+3=0
sinx=-3/2 (reject, (-1< sinx <1)
..
ii) 2 sin.x + 3 cos.x = 0
2sinx=-3cosx
divide by cosx
2sinx/cosx=-3
sinx/cosx=-3/2
tanx=-3/2
x=123.7º and 303.7º (quadrants II & IV)(reference angle=56.3º