SOLUTION: solve each equation. assume 0<=t<=2pi and 0<=theta<=360degrees a. sin(t)=1/2 b. 2cos^2 theta - 3cos theta +1=0 c. sec theta= -5 give answer accurate to 3 deci

Algebra ->  Trigonometry-basics -> SOLUTION: solve each equation. assume 0<=t<=2pi and 0<=theta<=360degrees a. sin(t)=1/2 b. 2cos^2 theta - 3cos theta +1=0 c. sec theta= -5 give answer accurate to 3 deci      Log On


   

Question 101198: solve each equation.
assume 0<=t<=2pi and 0<=theta<=360degrees
a. sin(t)=1/2

b. 2cos^2 theta - 3cos theta +1=0

c. sec theta= -5
give answer accurate to 3 decimals

d. 2 cos(2t)=7cos(t)
give answer accurate to 3 decimals
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
a. sin(t) = 1/2
t = pi/6
t = pi/6 and t = 5*pi/6
graph%28700%2C200%2C-1%2C6.3%2C-3%2C3%2C1%2F2%2Csin%28x%29%29
b. 2*cos^2(theta) - 3*cos(theta) + 1 = 0
2*cos^2(theta) - 2*cos(theta) - cos(theta) + 1 = 0
2*cos(theta)(cos(theta) - 1) - 1(cos(theta) - 1) = 0
(2*cos(theta) - 1)(cos(theta) - 1) = 0
2*cos(theta) - 1 = 0 and cos(theta) - 1 = 0
cos(theta) = 1/2 and cos(theta) = 1
theta = pi/3, 5*pi/3, 0, and 2*pi
graph%28700%2C200%2C-1%2C6.3%2C-1%2C1%2C2%2Acos%28x%29%5E2+-+3%2Acos%28x%29+%2B+1%29
c. simple
d. 2*cos(2t) = 7cos(t)
2*cos(2t) - 7*cos(t) = 0
2( cos(t + t) ) - 7*cos(t) = 0
2( cos(t)^2 - sin(t)^2 ) - 7*cos(t) = 0
2( cos(t)^2 - ( 1 - cos(t)^2 )) - 7*cos(t) = 0
2( 2*cos(t)^2 - 1) - 7*cos(t) = 0
4*cos(t)^2 - 7*cos(t) - 2 = 0
4*cos(t)^2 - 8*cos(t) + cos(t) - 2 = 0
4*cos(t)(cos(t) - 2) + 1(cos(t) - 2) = 0
(4*cos(t) + 1)(cos(t) - 2) = 0
4*cos(t) + 1 = 0 and cos(t) - 2 = 0
cos(t) = -1/4 and cos(t) = 2 ... nonexistant
I do not have a calculator with the inverse process with me ...