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## Wednesday, 30 July 2014

### CHAPTER 7- TRIGONOMETRIC EQUATIONS

Some equations which involve trigonometric functions of the unknown may be readily solved by using simple algebraic ideas (as Equation 1 below), while others may be impossible to solve exactly, only approximately (e.g., Equation 2 below):

EXAMPLE 1: Find all solutions of the equation  .
Solution: We can graphically visualize all the angles u which satisfy the equation by noticing that  is the y-coordinate of the point where the terminal side of the angle u (in standard position) intersects the unit circle (see Figure 1):
We can see that there are two angles in  that satisfy the equation:  and  . Since the period of the sine function is  , it follows that all solutions of the original equation are:

Find all solutions of the equation  .
Solution: Let u=2A; the equation is then equivalent to  , for which the solutions are (see EXAMPLE 1):

Hence the solutions for A are:

Find all solutions of the equation  that lie in the interval
Solution: The left hand side of the equation can be factored as:

hence either  or  . For  ,

while

and

The solution set of the original equation is then

Solve the equation  . Restrict solutions to the interval  .
Solution: The substitution  yields the equation  , which is quadratic in u. We use the quadratic formula to solve for  :

If  the calculator gives x=1.1191 as the acute solution, so we deduce that the other solution is  . The equation  has no solutions, since  for all x. The solution set is therefore  .