Find all angles, 0, degrees, is less than or equal to, A, is less than, 360, degrees0 ∘ ≤A<360 ∘ , that satisfy the equation below, to the nearest tenth of a degree (if necessary).7, cosine, A, minus, 2, equals, cosine, A, plus, 47cosA−2=cosA+4
Question
Find all angles, 0, degrees, is less than or equal to, A, is less than, 360, degrees0 ∘ ≤A<360 ∘ , that satisfy the equation below, to the nearest tenth of a degree (if necessary).7, cosine, A, minus, 2, equals, cosine, A, plus, 47cosA−2=cosA+4
Solution 1
The given equation is 7cosA - 2 = cosA + 4.
First, let's simplify this equation by subtracting cosA from both sides:
7cosA - cosA = 2 + 4
This simplifies to:
6cosA = 6
Next, divide both sides by 6:
cosA = 1
The cosine function equals 1 at two points in the interval 0 ≤ A < 360: at 0 degrees and 360 degrees. However, since 360 degrees is not included in the interval, the only solution in this interval is A = 0 degrees.
Solution 2
The given equation is cosA - 2 = cosA + 4.
Step 1: Subtract cosA from both sides of the equation to isolate the constants. This gives us -2 - cosA = 4.
Step 2: Simplify the equation by multiplying through by -1. This gives us 2 + cosA = -4.
Step 3: Subtract 2 from both sides of the equation to isolate cosA. This gives us cosA = -6.
However, the range of the cosine function is -1 ≤ cosA ≤ 1. Therefore, there are no angles A that satisfy the equation cosA = -6.
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