Question 1138070: solve 4 cos^2 θ + 2 sin θ = 3 for 0° ≤ θ ≤ 360°
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your equation is:
4 * cos^2(t) + 2 * sin(t) = 3
since cos^2(t) is equal to 1 - sin^2(t), this equation can be shown as:
4 * (1 - sin^2(t)) + 2 * sin(t) = 3
simplify to get:
4 - 4 * sin^2(t) + 2 * sin(t) = 3
subtract 3 from both sides of this equation to get:
-4 * sin^2(t) + 2 * sin(t) + 1 = 0
let x = sin(t) and this equation becomes:
-4 * x^2 + 2 * x + 1 = 0
factor this quadratic equation to get:
x = 0.80901699437495 or x = -0.30901699437495
since x = sin(t), this means that sin(t) = 0.80901699437495 or sin(t) = -0.30901699437495
when sin(t) = 0.80901699437495, your angle is either 54 degrees is 126 degrees.
when sin(t) = -0.30901699437495, your angle is either 198 degrees or 342 degrees.
your solution should therefore be either 54, 126, 198, or 342 degrees in the 0 to 360 degree interval.
you would confirm by evaluating the original equation at each of those angles.
if you do so, you will see that the original equation is true for all of them.
therefore, your solution is 54, 126, 198, and 342 degrees in the 0 to 360 degree interval.
this can be seen graphically as shown below:
in this graph, you set up two equations.
the first is y = 4 * cos^2(x) + 2 * sin(x)
the secon is y = 3
if the first equation is equal to the second equation, the graphs ill intersect and the intersection points will be as shown.
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