SOLUTION: In studying the emission of light, in order to determine the angle at which the intensity is a giving value, the equation sin^2(A) - 4sin(A) + 1 = 0 must be solved. Find the angl

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Question 157260: In studying the emission of light, in order to determine the angle at which the intensity is a giving value, the equation sin^2(A) - 4sin(A) + 1 = 0 must be solved. Find the angle A (to 0.1 of a degree). (sin^2(A) = (sinA)^2.
Answer by Fombitz(13823) About Me  (Show Source):
You can put this solution on YOUR website!
sin%5E2%28A%29-4sin%28A%29%2B1=0
Looks like a quadratic equation in disguise.
Let's substitute,
u=sin%28A%29
Then
sin%5E2%28A%29-4sin%28A%29%2B1=0
u%5E2-4u%2B1=0
Using the quadratic formula, solve for u.
u+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
u+=+%28-%28-4%29+%2B-+sqrt%28+%28-4%29%5E2-4%2A1%2A1+%29%29%2F%282%2A1%29+
u+=+%284+%2B-+sqrt%28+16-4+%29%29%2F%282%29+
u+=+%284+%2B-+sqrt%28+12%29%29%2F%282%29+
Two solutions,
u%5B1%5D+=+%284+%2B+sqrt%28+12%29%29%2F%282%29+
u%5B1%5D+=+%284+%2B+3.46%29%2F%282%29+
u%5B1%5D+=+%287.46%29%2F%282%29+
u%5B1%5D+=+3.73+
Substitute u=sin%28A%29.
u%5B1%5D+=+3.73+
sin%28A%29+=+3.73+
We throw this solution out since abs%28sin%28A%29%29%3C=1
u%5B2%5D=%284+-+sqrt%28+12%29%29%2F%282%29+
u%5B2%5D=%284+-+3.46%29%2F%282%29+
u%5B2%5D=%280.54%29%2F%282%29+
u%5B2%5D=0.27
Substitute u=sin%28A%29.
u%5B2%5D=0.27
sin%28A%29=0.27
A=+sin%5E%28-1%29%280.27%29+
A=15.6
and by identity,
A=180-15.6=164.4