Hello :
by idetity : 1+cos(2a) = 2cos²(a)
let : a = <span>θ/2
1+cos(2×</span>θ/2)=2cos²(θ/2)
1+cos(θ) = 2cos²(θ/2)
solve : 2 cos^2 (θ/2) - 3 cos(θ) = 0 ......(1)
(1) : 1+cos(θ)-3 cos(θ) =0
2cos(θ) = 1
cos(θ) = 1/2
in : [0, 2π<span>[
</span>( θ = π/3) or (θ = 5π/3)
Answer: -6 +/-
= x
<u>Step-by-step explanation:</u>
y = x² + 12x - 5
<u> +5 </u> <u> +5 </u>
y + 5 = x² + 12x
<u> + 36 </u> <u> +36 </u> <em>added (12 ÷ 2)² to both sides</em>
y + 41 = (x + 6)²
To find the zeros, let y = 0:
0 + 41 = (x + 6)²

= x + 6
<u>-6 </u> <u> -6 </u>
-6 +/-
= x
Answer:
Step-by-step explanation:
Hey remember it me again
Answer:
One number that works is 1.
Step-by-step explanation:
(1-2)(1+7)
-1(1+7)
-1*8
-8
Answer:
You know you got one solution if when you solve your equation its not for example 5=5 because that's infinite solutions or 5=6 because that's no solution so one solution would be like y=5