Solve.........?

Question

Solve sin x + 1 = 0 for 0°≥ x ≥ 360°

I'm having a lot of trouble with this.. I need to understand it for a test.

Any help will be appreciated.

Answer 1

sin x + 1 = 0 <=> sin x = -1
The only x with 0° ≤ x ≤ 360° for which sin x = -1 is x = 270° - the unit circle should show you this.

Note that I assume it is 0° ≤ x ≤ 360° and not 0° ≥ x ≥ 360° as you wrote, since there's no x for which 0° ≥ x ≥ 360°.

Answer 2

You need to find where sin x is equal to -1 so that you will be left with -1+1=0.
Between 0 and 360, the only place where sin x equals -1 is 270.
If you know your unit circle, you can see where the y value is -1, an that is sin x.

Answer 3

Recall that the graph of sin x is one sinusoidal curve between zero and 350 degrees.

Taking that into consideration,
sin x + 1 = 0
sin x = -1

Being reminded of the graph, you can get

x= 270 degrees

or

consider x = 1

the basic angle is 90 degrees

add 180 degrees you get 270

this is so as we follow the ASTC idea where sin is negative in the third and fourth quadrant. Only 270 can be somewhere there when you try to solve it.

Hope this helps

Answer 4

sin x + 1= 0

sin x = -1

between 0 and 360, if you check the circle you can see

x= 270 oC

http://tanopah.jo.free.fr/seconde/trig2248.gif

Answer 5

sinx=-1
x=270'
or x=3pi/2 rad

Answer 6

i got x= -89.9995