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Trigonometric EquationsTrigonometric Identities : Double Angles : Trigonometric Graphs : Equations 
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In this essay we will combine the Trigonometric Function into equations that can be solved.
We begin by reminding ourselves of the trigonometric relations:
In addition, there are relations called double angles:
Because Sin^{2}X + Cos^{2}X = 1, this last relation can also be written as:
Sines are periodic. They oscilate between 1 and 1 over 360^{o} (2 π Radians) begining and ending at 0.
Below is the graph of Y = Cos X. This is similar but in a different phase.
Cosines also oscilate between 1 and 1 over 360^{o} (2 π Radians) begining and ending with 1.
The table below summarises the information for both Sines and Cosines between 0^{o} and 360^{o} (0 to 2 π Radians). This information will be used when solving trigonometric equations.
Angle (^{o}) 
Angle (Rad) 
Sine  Cosine 

Sin X = 1 / 2
Using the table, it is easy to see that X has two values in the required range. They are:
X = 30^{o} and X = 150^{o}.
Cos X + 1 / 2 = 0
Rearranging the equation (to get Cos X on one side and the numbers on the other side) gives:
Cos X = 1 / 2
Using the table, we can see that X has two values in the required range. They are:
X = 120^{o} and X = 240^{o}.
Cos X Tan X = 1 / √2
Using the identity to replace Tan X gives:
Cos X (Sin X / Cos X) = 1 / √2
The Cosines cancel out to give:
Sin X = 1 / √2
This gives two values of X:
X = 45^{o} and X = 135^{o}.
2 Cos 2X + 1 = 0
Rearrange the equation:
Cos 2X =  1 / 2
Therefore 2X = 120^{o} and 240^{o} which gives:
X = 60^{o} and X = 120^{o}.
Sin X  Cos 2X = 0
Using the double angle identity, replace Cos 2X by (1  2 Sin^{2}X):
Sin X  (1  2 Sin^{2}X) = 0
Sin X  1 + 2 Sin^{2}X = 0
which rearranges to a quadratic equation in Sin^{2}X:
2 Sin^{2}X + Sin X  1 = 0
This can be solved by factorising:
(2 Sin X  1)(Sin X + 1) = 0
This equation gives 0 if either 2 Sin X  1 = 0 or Sin X + 1 = 0. In other words:
Sin X = 1 / 2 or Sin X = 1
The first equation gives two vales (X = 30^{o}, X = 150^{o}), the second equation gives one value (270^{o}). Thus the solution of the original equation is:
X = 30^{o}, X = 150^{o} and X = 270^{o}.
2 Cos^{2}X + Sin 2X = 0
Using the double angle identity, the Sin 2X can be replaced by 2 Sin X Cos X:
2 Cos^{2}X + 2 Sin X Cos X = 0
2 Cos X is common to both terms so this can be rewritten:
2 Cos X ( Cos X + Sin X) = 0
This equation gives 0 if either 2 Cos X = 0 or Cos X + Sin X = 0. In other words:
Cos X = 0 or Sin X = Cos X
The first equation gives two vales (X = 90^{o}, X = 270^{o}). The second equation also gives two values (135^{o} and 315^{o}  check these figures in the table). Thus the solution of the original equation is:
X = 90^{o}, X = 135^{o}, X = 270^{o} and 315^{o}.
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