A few days ago
Trig—How do I solve for angle measure x?
I need to solve for x for both of these equations.
sin(x)- (square root of 3)cos(x) = 1
and
3(-cos(x)) = sin^2(x)
help please?
Top 1 Answers
A few days ago
Favorite Answer
sin(x) – sqrt(3)cos(x) = 1
Divide by sqrt(1 + 3) = 2:
(1/2)sin(x) – (sqrt(3)/2)cos(x) = 1/2
cos(pi/3)sin(x) – sin(pi/3)cos(x)= 1/2
sin(x – pi/3) = 1/2
sin(x – pi/3) = sin(pi/6)
x – pi/3 = 2n pi + pi/6 or x = 2n pi + 5pi/6 for any integer n
x = 2n pi + pi/2
or
x = 2n pi + 7pi/6.
-3cos(x) = sin^2(x)
– 3cos(x) = 1 – cos^2(x)
cos^2(x) – 3cos(x) – 1 = 0
cos(x) = ( 3 +/- sqrt(9 + 4) ) / 2
cos(x) = (3 +/- sqrt(13) ) / 2
As (3 + sqrt(13) ) / 2 > 1,
x = 2n pi +/- arccos( (3 – sqrt(13)) / 2 ) for any integer n.
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