Test Corrections: solve tan(15x) = -1, sec(2) = sqrt 2, etc

[Lokito]

I did badly on a trig test. I can do corrects and re-take the test, but I haven't figured out some of the answers.

I need help with the last problem of the second section.
16). Solve the equations for 0° (less than or equal to) x (less than or equal to) 360°
A) tan(15x) = -1 (Directions: DO NOT find all 30 solutions, only find the first two)
tan^-1(-1) = -45
15x = -45
x1 = -3
How do I solve for the second solution?

B) sec(x) = sqrt 2
Help?

What I've done for the calculator section of the test:

1). Find the coordinates (both x and y) of a point that lies at an angle of 4.5 radians on a circle with radius 8.

8cos(4.5)= -1.68

Do I need more than one answer?

4). Use the Pythagorean Identities or triangles to solve the following problem.
cos(x)=-.482 and (pi)/2 (less than or equal to) x (less than or equal to) pi
Find sin and tan.

sin^2(x) = cos^2(x) = 1, etcetera.
Sin(x) = .877
Tan(x) = sin(x) / cos(x) = -1.82

5). Solve the following equation for 0(less than or equal to) x (less than or equal to) 2Pi

8sin(1/2x)-4=3

8sin(1/2x)=7
sin(1/2x)=7/8
sin^-1(7/8)=1.065
1.065 * 2 = 2.131 = x1
x2 = 4.152

Solve the following equation for 0° (less than or equal to) x (less than or equal to) 360°
3cos(2x)=-2

cos(2x)=-2/3
cos^-1(-2/3)= 131.81
131.81 / 2 = 65.905° = x1
x2 = 114.09°
x3 = 275.91°
x4 = 294.09°

NO CALCULATOR SECTION

11). Give exact answers for the following:

a) sin(135°) = sqrt2 / 2
b) cos (7Pi / 6) = - sqrt3 / 2
c) cos (-2Pi / 3) = -1/2
d) tan(2Pi / 3) = 1/2
f) tan(45°) = 1
g) cos(8Pi / 3) = cos(480°) = cos(120°) = -1/2

12) b) Derive the other two Pythagorean Identities from the first one.
(know how to do this, blanked on the test)

13). a) Convert 18° to radians
18° * (Pi/180) = Pi/10

b) Convert 4Pi / 15 to degrees
4Pi/15 * 180/Pi = 48°

15). Solve the following equations for ...

b) cos(2x) = 1/2
2x=60 or Pi/3
x1 = Pi/6
x2 = 5/6 Pi
x3 = 7/6 Pi
x4 = 11/6 Pi

 

[masters]

B sec x = sqrt 2

sec x = 1/cos x

1/cos x = sqrt 2

cos x = 1/sqrt 2

cos[sup:1e2yy9i8]-1[/sup:1e2yy9i8] (1/sqrt 2) = 45 degrees

 

[galactus]

I need help with the last problem of the second section.
16). Solve the equations for 0° (less than or equal to) x (less than or equal to) 360°
A) tan(15x) = -1 (Directions: DO NOT find all 30 solutions, only find the first two)
tan^-1(-1) = -45
15x = -45
x1 = -3

Note that they only what the solutions between 0 and 360, so -3 is not what you want.
I am going to use radians.

You get -3, which is \(\displaystyle \frac{-\pi}{60}\). Now add \(\displaystyle \frac{\pi}{15}\) to get your first solution, then add it again to find your second solution.

If you want to remain in degrees, then add 12 degrees consecutively. Because 12 degrees is \(\displaystyle \frac{\pi}{15}\) radians.