If you would like an review of trigonometry, click on trigonometry.
Solve for x in the following equation.
There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.
We know that the e function is positive in the first and the second
quadrant. Therefore two of the solutions are the angle 5x that terminates
in the first quadrant and the angle
that terminates in the second
quadrant. We have already solved for 5x.
The exact solutions are and
The period of the sin function is This means that the values will repeat every radians in both directions. Therefore, the exact solutions are and where n is an integer.
The approximate solutions are and where n is an integer.
These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.
Check the answer x=0.028669513781
Since the left side equals the right side when you substitute 0.028669513781 for x, then 0.028669513781 is a solution.
Check the answer x=0.599649016937
Since the left side equals the right side when you substitute for x, then is a solution.
Graph the equation
Note that the graph crosses the x-axis many times indicating many solutions.
Note that it crosses at 0.028669513781. Since the period is , it crosses again at 0.028669513781+1.256637=1.285306and at 0.028669513781+2(1.256637)=2.54194, etc.
The graph crosses at 0.599649016937. Since the period is , it will cross again at and at 0.599649016937+2(1.256637)=3.112923, etc
If you would like to test yourself by working some problems similar to this example, click on Problem.
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