2 cos2 6θ + 3 cos 6θ + 1 = 0
(2cos6θ+1)(cos6θ+1) = 0
cos6θ = -1/2 or -1
so,
6θ = 2π/3 or 4π/3
θ = π/9+kπ/3 or 2π/9+kπ/3, k=0..5
6θ = π
θ = π+kπ/3, k=0..5
You add kπ/3 since the solutions are (6θ+2π)k
Find all degree solutions. (Enter your answers as a comma-separated list. Let k be any integer.)
2 cos2 6θ + 3 cos 6θ + 1 = 0
6 answers
how would I convert these to degrees?
why would you want to?
Ok, if you must, recall that π = 180°
Ok, if you must, recall that π = 180°
Thank you. The answer came out to
20degree + 60degreeK, 30degree + 60 degreeK, 40degree + 60degreek.
20degree + 60degreeK, 30degree + 60 degreeK, 40degree + 60degreek.
right, but I'd write it as
(20+60k)°, (30+60k)°, (40+60k)°
I see you caught my typo.
(20+60k)°, (30+60k)°, (40+60k)°
I see you caught my typo.
Solve the equation for all degree solutions and if
0° ≤ θ < 360°.
Do not use a calculator. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
2 sin θ −
3
= 0
(a) all degree solutions (Let k be any integer.)
θ =
(b) 0° ≤ θ < 360°
θ =
0° ≤ θ < 360°.
Do not use a calculator. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
2 sin θ −
3
= 0
(a) all degree solutions (Let k be any integer.)
θ =
(b) 0° ≤ θ < 360°
θ =