<< Chapter < Page | Chapter >> Page > |
Solve the equation exactly:
Using grouping, this quadratic can be factored. Either make the real substitution, or imagine it, as we factor:
Now set each factor equal to zero.
Next solve for as the range of the sine function is However, giving the solution
Solve [Hint: Make a substitution to express the equation only in terms of cosine.]
Solve exactly:
This problem should appear familiar as it is similar to a quadratic. Let The equation becomes We begin by factoring:
Set each factor equal to zero.
Then, substitute back into the equation the original expression for Thus,
The solutions within the domain are
If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.
Solve the equation quadratic in form exactly:
We can factor using grouping. Solution values of can be found on the unit circle:
While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.
Use identities to solve exactly the trigonometric equation over the interval
Notice that the left side of the equation is the difference formula for cosine.
From the unit circle in [link] , we see that when
Solve the equation exactly using a double-angle formula:
We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:
So, if then and if then
Notification Switch
Would you like to follow the 'Precalculus' conversation and receive update notifications?