Find all angles, 0, degrees, is less than or equal to, B, is less than, 360, degrees0 ∘ ≤B<360 ∘ , that satisfy the equation below, to the nearest tenth of a degree (if necessary).12, cosine, B, plus, 2, equals, 4, cosine, B, plus, 112cosB+2=4cosB+1
Question
Find all angles, 0, degrees, is less than or equal to, B, is less than, 360, degrees0 ∘ ≤B<360 ∘ , that satisfy the equation below, to the nearest tenth of a degree (if necessary).12, cosine, B, plus, 2, equals, 4, cosine, B, plus, 112cosB+2=4cosB+1
Solution
The given equation is 12cosB + 2 = 4cosB + 1.
First, let's simplify this equation by subtracting 4cosB from both sides:
12cosB - 4cosB = 1 - 2
This simplifies to:
8cosB = -1
Next, divide both sides by 8 to isolate cosB:
cosB = -1/8
Now, we need to find the angle B that satisfies this equation. We can do this by taking the inverse cosine (or arccos) of both sides:
B = arccos(-1/8)
Using a calculator, we find that the principal value of B is approximately 173.6 degrees.
However, since the cosine function is negative in both the second and third quadrants, there is another solution in the range 0 ≤ B < 360. The solution in the third quadrant is 180 + (180 - 173.6) = 186.4 degrees.
So, the solutions to the equation are B = 173.6 degrees and B = 186.4 degrees.
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