8sinA = 4 + cosA then sinA =

8sinA = 4 + cosA then sinA = ?

sinA=2t/(1+t^2) and cosA=(1-t^2)/(1+t^2), where t=tan(A/2).
This 16t/(1+t^2) = 4+(1-t^2)/(1+t^2) and after some fumbling you get
3t^2 -16t+5=0 giving t=1/3 or t=5.
giving sinA= 3/5 or 5/13.

An alternative method is to express cosA=8sinA-4 so cos^2(A)=64sin^2(A)-64sinA+16
but cos^2(A)=1-sin^2(A) giving 65sin^2(A)-64sinA+15=0.
Solving for sinA gives sinA=5/13 or 3/5 as above. However, you need to
verify that these really are solutions since cosA has been squared which could
introduce extraneous solutions.

Rewrite the equation as 8*sin(A) - 4 = cos(A) and then square both sides to get

64*sin^2(A) - 64*sin(A) + 16 = cos^2(A) = 1 - sin^2(A) --->

65*sin^2(A) - 64*sin(A) + 15 = 0, which is quadratic in sin(A), so using the QF

sin(A) = (64 +/- sqrt[(-64)^2 - 4*65*15]) / (2*65) = (64 +/- 14)/130.

Thus sin(A) = 50/130 = 5/13 or sin(A) = 78/130 = 39/65 = 3/5.

Using the 'square both sides' method at the beginning can introduce extraneous
solutions sometimes, so I'm just going to double-check here. There are also
some issues regarding which quadrant angle A is in. For instance, if A is in
quadrant ll and sin(A) = 5/13, then cos(A) = -12/13 and the equation is
satisfied, but not if A was in any other quadrant. If A is in quadrant l then
if sin(A) = 3/5 we get cos(A) = 4/5 and the equation is satisfied, but not if
A is in any other quadrant.
Thus, sin(A) = 5/13 if A is in quadrant ll and sin(A) = 3/5 if A is in quadrant l.

Given:

8sin(A) = 4 + cos(A)

8sin(A) - 4 = cos(A)

64sin²(A) - 32sin(A) - 32sin(A) + 16 = cos²(A)

64sin²(A) - 64sin(A) + 16 = 1 - sin²(A)

65sin²(A) - 64sin(A) + 15 = 0

Using the quadratic formula,

sin(A) = 5/13 and sin(A) = 78/130