Solve 4cos^2A = 3cosA for 90º≤A≤180º. (Enter only the number.)
Find A when 0º≤A≤90º and 3tan^2A = 2tanA + 1. (Enter only the number.)
+33661 1. If 4cos2A = 3cosA then one solution is cosA = 0 which occurs when A =90°
Another solution is when cosA ≠ 0, then you can cancel a cosA from both sides leaving 4cosA = 3 or cosA = 3/4. However, in the range 90°≤A≤180° cosA is 0 or negative, so cannot be 3/4. Therefore the only solution is A = 90°
2. Rewrite the second equation as 3tan2A - 2tanA -1 = 0
This can be factored as (3tanA + 1)*(tanA - 1) = 0 so either tanA = -1/3 or tanA = 1
In the range 0°≤A≤90° tanA must be positive, so the only solution is tanA = 1 or A = tan-1(1) or A = 45°
.
+33661 1. If 4cos2A = 3cosA then one solution is cosA = 0 which occurs when A =90°
Another solution is when cosA ≠ 0, then you can cancel a cosA from both sides leaving 4cosA = 3 or cosA = 3/4. However, in the range 90°≤A≤180° cosA is 0 or negative, so cannot be 3/4. Therefore the only solution is A = 90°
2. Rewrite the second equation as 3tan2A - 2tanA -1 = 0
This can be factored as (3tanA + 1)*(tanA - 1) = 0 so either tanA = -1/3 or tanA = 1
In the range 0°≤A≤90° tanA must be positive, so the only solution is tanA = 1 or A = tan-1(1) or A = 45°
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