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

Anonymous

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

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.

Principles of Mangement?

give me information about empower others?

does anybody know what the latin for panda is. thanks.?

Can someone help me with this lol?

whats a good attention getter for my essay?

information on Passchendaele?