Solve 5 sin A - cos A = 0, for 0 ≤ A < 2π (round to the nearest hundredth of a radian).

Question

I have tried my hardest to answer this question correctly, but it is always in correct. Thanks for the help! (Solve 5 sin A - cos A = 0, for 0 ≤ A ≤ 2π (round to the nearest hundredth of a radian).

Answer 1

To solve this problem, you need to get the functions all the same (instead of a sine and a cosine. How about if we divide everything by cos(A)? That will convert the sin(A) into a tan(A) and it will convert cos(A) into 1. The only thing we need to worry about is the case when cos(A) = 0 because we might lose an answer. But cos(A) is 0 only at A = pi/2 and 3pi/2 and that sin(A) will be 1 or -1 and that does not result in an answer so we are OK to divide by cos(A).

So:

5sin(A) - cos(A) = 0

5sin(A)/cos(A) - cos(A)/cos(A) = 0/cos(A)

5tan(A) - 1 = 0

5tan(A) = 1

tan(A) = 1/5

A = tan^-1(1/5)

Using a calculator, A = 0.197396

Looking at the angle on the coordinate plane, we see that tan can also be 1/5 in quadrant 3:

So I also need to consider the answer where A = 0.197396 + pi = 3.338988

So A = 0.20 radians and 3.34 radians

Answer 2

5 sin A - cos A = 0

5 sin A = cos A

5 = cos A / sin A

1/5 = sin A / cos A

0.2 = tan A

0.19739 = A

3.33898 = A