Trig Equations: Solving for x in Given Ranges

[xdeanna]

I had some questions on a test and i got these two wrong..

x is greater than or equal to 0 and x is less than or equal to 360

1) 2sin x + sin x cos x = 0
2) sin 2x = -1/sqrt(2)

 

[Mark44]

I had some questions on a test and i got these two wrong..

x is greater than or equal to 0 and x is less than or equal to 360

1) 2sin x + sin x cos x = 0
2) sin 2x = -1/sqrt(2)

Show us what you did and we'll go from there...

 

[xdeanna]

2Sin x + cos x sin x = 0
Sin x(2 + cos x) = 0
sin x = 0
x= sin^-1(0)
x=0degrees
or
2+ cos x= 0
cos x =-2
x= cos^-1(-2)
x=?

I did something different on the test but i think this is wrong too >.<

Don't make show what I did for #2 :(

 

[praharmitra]

Well, cos x = 2 does not have any solutions in x. This is because the range of cos x is [-1, 1]

 

[Mark44]

2Sin x + cos x sin x = 0
Sin x(2 + cos x) = 0
sin x = 0
x= sin^-1(0)
x=0degrees

This is one solutions; there are two other numbers x in [0, 360] (deg.) for which sin(x) = 0. The inverse sine function will only produce numbers in the interval [-90, 90], in degrees. Think about where the graph of y = sin(x) crosses the x-axis.

or
2+ cos x= 0
cos x =-2
x= cos^-1(-2)
x=?

I did something different on the test but i think this is wrong too >.<

Don't make show what I did for #2 :(

OK, maybe I can help you get started in the right direction. If sin(2x) = -1/sqrt(2), then 2x = sin^-1(-1/sqrt(2)). What are the possible values of 2x in [0, 720] (deg), so that sin(2x) = -1/sqrt(2)?

 

[snshusat161]

it's long time that I haven't posted anything here and so I forgot everything about latex. Will anybody here consider giving me link so that I can learn to type it.

then I'll post solution for this problem.