Solve the following equations for values of q in the range:
0 < q < 2p.
1) 2 cos q - 1 = 0
Solution:
Transpose to get angle on one side:
2 cos q = 1
cos q = 1/2
q = p/ 3 (From table)
And: q = 5p/ 3 for next quadrant, i.e.
2) Ö3 tan2 q + 2tan q - Ö3 = 0
Solution:
Rewrite as quadratic eqn. in x:
Ö3 x2 + 2 x - Ö3 = 0
Solve using the quadratic formula:
a =Ö3, b = 2, c = -Ö3 we have:
b 2 - 4ac = 4 - 4(-3) = 16
Substituting values for the full formula:
x = - 2 + Ö16/ 2Ö3 = - 2 +4 / 2Ö3 = -6/ 2Ö3 , 2/ 2Ö3 ,
= -3/ Ö3 , 1/Ö3 ,
x = - Ö3, x = Ö3/3
Substituting back angle functions:
tan q = - Ö3, tan q = Ö3/3
For: tan q = Ö3/3, q = p/ 6, 7p/ 6
For: tan q = -Ö3, q = - p/ 3 (not in range), 2p/ 3, 5p/ 3
So angle solns over range 0 < q < 2p:
p/ 6, 7p/ 6, 2p/ 3, 5p/ 3
3) 2Ö3 csc 2 q - 7 csc q + 2Ö3 = 0
Solution:
Rewrite as quadratic eqn. in x:
2Ö3 x2 - 7 x + 2Ö3 = 0
Solve for x using the quadratic formula:
a =2Ö3, b = - 7, c = 2Ö3 we have:
b 2 - 4ac = 49 - 4(12) = 1
Substituting values for the full formula:
x = 7 + 1/4Ö3 = 8/4Ö3 , 6 /4Ö3
= 2/Ö3, 3/ 2/Ö3 = 2Ö3/3, Ö3/2
Substituting back angle functions:
For: csc q = 2Ö3/3, q = p/ 3 (from table)
And also:
For: csc q = Ö3/2 = 0.866, q does not exist
Rem: csc q = 1/ sin q so sin q = 1/ 0.866 =1.1547
Which is impossible, i.e. NO sin q >1
4) sec 2 q - 3 sec q + 2 = 0
Solution: Factor as a regular quadratic eqn.
(sec q - 2) (sec q - 1) = 0
sec q - 2 = 0
sec q = 2
q = p/ 3 (From table)
Also include: q = 5p/ 3 , 7p/ 3
For 2nd factor:
sec q - 1 = 0
sec q =1
q = 0 0
5) sec 3 q + 2 = 0
Solution:
Transpose sec to one side:
sec 3 q = - 2
Convert to cosine function and take reciprocal of both sides:
(Since: sec q = 1/ cos q)
Then: cos 3 q = - ½
This function is -ve in quadrants 2, 3 so:
3q = 2p/ 3, 4p/ 3,
From here other q solns. can also be found as per the method shown in Part 1, ex. 4, i.e.
The table below was used to obtain the angles from the trig functions:
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