Chapter 1 | Functions and Graphs
69
Solution a. Using the double-angle formula for cos(2 θ ), we see that θ is a solution of 1+cos(2 θ ) =cos θ
if and only if
1+2cos 2 θ −1=cos θ ,
which is true if and only if
2cos 2 θ −cos θ =0. To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by cos θ . The problem with dividing by cos θ is that it is possible that cos θ is zero. In fact, if we did divide both sides of the equation by cos θ , we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that θ is a solution of this equation if and only if cos θ (2cos θ −1) =0.
Since cos θ =0 when
θ = π
π 2 ±
π , π
π ,…,
2 ,
2 ±2
and cos θ =1/2 when
θ = π
π 3 ±2
π ,…or θ = − π
π 3 ±2
π ,…,
3 ,
3 , −
we conclude that the set of solutions to this equation is θ = π 2 + nπ , θ = π 3 +2 nπ , and θ = − π 3 +2 nπ , n =0, ±1, ±2,…. b. Using the double-angle formula for sin(2 θ ) and the reciprocal identity for tan( θ ), the equation can be written as 2sin θ cos θ = sin θ cos θ . To solve this equation, we multiply both sides by cos θ to eliminate the denominator, and say that if θ satisfies this equation, then θ satisfies the equation 2sin θ cos 2 θ −sin θ =0. However, we need to be a little careful here. Even if θ satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by cos θ . However, if cos θ =0, we cannot divide both sides of the equation by cos θ . Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor sin θ out of both terms on the left-hand side instead of dividing both sides of the equation by sin θ . Factoring the left-hand side of the equation, we can rewrite this equation as sin θ (2cos 2 θ −1) =0.
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