you know that the reference angle is θ = π/3, right? Because cos π/3 = 1/2
Now, since cosπ = x/r, you need x negative, so you are in QII and QIII.
So, draw your triangle on the negative x-axis, giving you π-π/3 and π+π/3 = 2π/3 and 4π/3.
Since cosine has a period of 2π, you can add or subtract 2π as many times as you like and get the same value for cosθ
That gives you the answer in part 1.
Now, having those solutions, just plug in different values of n:
n = 1: 2π/3 + 2π*1 = 8π/3
n = -3: 4π/3 + 2π(-3) = -14π/3
and so on for as many values as you need.
Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.)
cos θ = −1/2
1).θ= Answer-->(2π/3)+2πn, 4π/3+2πn
2). List six specific solutions
a).
b).
c).
d).
e).
f).
I really need help getting these please. I've tried 3 times already.
1 answer
1 answer