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1. Solve the equation for -180<x<180

I remember the teacher saying something about cos(-x)=cos(x) but when I put the 2 in the calculator I get different answers.

What value do I use?
2. x+40=arcos(-0.5)
3. You don't need cos(-x)=cos(x) here, do you know how to solve things like cos(t)=0.5? If so then just do that with t=x+40.
4. So when I do cos-1(-0.5) and get 120 I use that as my initial angle?
I thought that the minus sign can. Be ignored and it just TELs you that the angle needs to be in the negative quadrant
(Original post by james22)
You don't need cos(-x)=cos(x) here, do you know how to solve things like cos(t)=0.5? If so then just do that with t=x+40.
5. (Original post by Merdan)
x+40=arcos(-0.5)
Not sure what arcos is
6. (Original post by techno-thriller)
Not sure what arcos is
arcos is inverse of cos, so cos^(-1) on your calculator.

It helps to either use the quadrant method or - the one I use - the graph method.

First, graph cos(x+40), or cosx might be easier. Then solve as simultaneous equations with y = cosx and y = -0.5. The points are the intersections/solutions. You should be able to use the symmetry of cosine to calculate the other answers within the range.

When you input inverse cosine of -0.5, you'll get one solution. This is 120 degrees, which is helpful, considering is it 30 degrees to the right of the first x intercept.
7. (Original post by Mike_123)
arcos is inverse of cos, so cos^(-1) on your calculator.

It helps to either use the quadrant method or - the one I use - the graph method.

First, graph cos(x+40), or cosx might be easier. Then solve as simultaneous equations with y = cosx and y = -0.5. The points are the intersections/solutions. You should be able to use the symmetry of cosine to calculate the other answers within the range.

When you input inverse cosine of -0.5, you'll get one solution. This is 120 degrees, which is helpful, considering is it 30 degrees to the right of the first x intercept.
Thanks for That.
8. (Original post by techno-thriller)
Solve the equation for -180<x<180

I remember the teacher saying something about cos(-x)=cos(x) but when I put the 2 in the calculator I get different answers.

What value do I use?
Ignore the negative sign at first and find arcos(0.5)=60
Then the negative sign tells us that our angle must be in the second or third quadrant so it is 120 or 240, but only 120 will give a solution in the required range.Does the attached file help?
Attached Images
9. Trig help.pdf (23.3 KB, 89 views)
10. (Original post by brianeverit)
Ignore the negative sign at first and find arcos(0.5)=60
Then the negative sign tells us that our angle must be in the second or third quadrant so it is 120 or 240, but only 120 will give a solution in the required range.Does the attached file help?
You took 60 from 180 because that's where cos is negative. Same with the addition. But looking at the graph, there are 3 intersections between y=cos x+40 and y=-0.5. How do you find the other values? Looking at the link, it explains it well but in this case we're going to -180, finding angles for 0<x<360 is pretty easy IMO. Thanks for the help.

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