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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

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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

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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.

This problem has been solved

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.

This problem has been solved

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