22Ftrigeqns

= Solving Trig Equations =

Because trig functions are cyclic (repeating over and over to infinity), any simple trig equation will have an infinite number of solutions.


 * Example 1 **

Solve: .. sin(q ) = 0.5

Graphically, the solution is the set of points where the curve ** y = sin( q ) ** intersects with the line ** y = 0.5 **.
 * Note: ** ** y = sin( q ) ** is positive in 1st and 2nd quadrants (Q1 and Q2) so intercepts occur in Q1 and Q2.

We can find one solution using the calculator. ... ... ... sin(q ) = 0.5 ... ... ... q = sin –1 (0.5) ... ... ... q = 30º ... ... ** {solution in Q1} **

We can find the solution in Q2 using ... ... ... q = 180º – 30º ... ... ... q = 150º .. ... ** {solution in Q2} **

Further solutions can be found by continuously adding 360º to these solutions: ... ... ... q = 30º ... ... ... q = 150º ... ... ... q = 30 + 360 ... ... so ... q = 390º ... ... ... q = 150 + 360 .... so ... q = 510º ... ... ... q = 390 +360 ... . so ... q = 750º ... ... ... <span style="font-family: Symbol,sans-serif; font-size: 110%;">q = 510 + 360 .... so ... <span style="font-family: Symbol,sans-serif; font-size: 110%;">q = 870º ... ... ... etc

We can also subtract 360º to get the negative solutions, so the full solution to sin(<span style="font-family: Symbol,sans-serif; font-size: 110%;">q ) = 0.5 is ... ... ... <span style="font-family: Symbol,sans-serif; font-size: 110%;">q = ..., –570º, –230º, –210º, 30º, 150º, 390º, 510º, 750º, ...

Often we are asked for solutions within a specified domain:
 * Note: **


 * Example 2 **

Solve: .. cos(<span style="font-family: Symbol,sans-serif; font-size: 110%;">q ) = –0.22 ... for <span style="font-family: Symbol,sans-serif; font-size: 110%;">q Î [0º, 360º] ... ... {this means find values of <span style="color: #008000; font-family: Symbol,sans-serif; font-size: 110%;">q between 0º and 360º}

cos(<span style="color: #000000; font-family: Symbol,sans-serif; font-size: 110%;">q ) is negative in 2nd and 3rd quadrants so solutions will be in Q2 and Q3}
 * Solution: **

{first find in Q1, solution to cos(a) = 0.22} ... ... ... cos(a) = 0.22 ... ... ... a = cos –1 (0.22) ... ... ... a = 77.3º

{to get solution in Q2, subtract from 180º} ... ... ... <span style="font-family: Symbol,sans-serif; font-size: 110%;">q = 180º – 77.3º ... ... ... <span style="font-family: Symbol,sans-serif; font-size: 110%;">q = 102.7º ... ... ** {solution in Q2} **

{to get solution in Q3, add to 180º} ... ... ... <span style="font-family: Symbol,sans-serif; font-size: 110%;">q = 180º + 77.3º ... ... ... <span style="font-family: Symbol,sans-serif; font-size: 110%;">q = 257.3º ... ... ** {solution in Q3} **

. ... .... <span style="font-family: Symbol,sans-serif; font-size: 110%;">q = 102.7º, 257.3º
 * Hence the solutions are **

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