College Trigonometry, 2011a

10.7 Trigonometric Equations and Inequalities 865
which lie in [0, 2π) arex = π 2 , 3π 2 , π 3 and 2π 3 . We graph y =sin(2x) andy = √ 3 cos(x) and,
after some careful zooming, verify our answers.
y =cos(3x) andy =cos(5x)
y =sin(2x) andy = √ 3 cos(x)
7. Unlike the previous problem, there seems to be no quick way to get the circular functions or
their arguments to match in the equation sin(x)cos ( ) (
x
2 + cos(x)sin x
)
2 =1. Ifwestareat
it long enough, however, we realize that the left hand side is the expanded form of the sum
formula for sin ( x + x ) (
2 . Hence, our original equation is equivalent to sin 3
2 x) = 1. Solving,
we find x = π 3 + 4π 3 k for integers k. Two of these solutions lie in [0, 2π): x = π 3 and x = 5π 3 .
Graphing y =sin(x)cos ( ) (
x
2 + cos(x)sin x
)
2 and y = 1 validates our solutions.
8. With the absence of double angles or squares, there doesn’t seem to be much we can do.
However, since the arguments of the cosine and sine are the same, we can rewrite the left
hand side of this equation as a sinusoid. 9 To fit f(x) = cos(x) − √ 3sin(x) to the form
A sin(ωt + φ)+B, we use what we learned in Example 10.5.3 and find A =2,B =0,ω =1
and φ = 5π 6 . Hence, we can rewrite the equation cos(x) − √ 3sin(x) =2as2sin ( x + 5π )
6 =2,
or sin ( x + 5π )
6 = 1. Solving the latter, we get x = −
π
3
+2πk for integers k. Only one of
these solutions, x = 5π 3
, which corresponds to k = 1, lies in [0, 2π). Geometrically, we see
that y = cos(x) − √ 3sin(x) andy = 2 intersect just once, supporting our answer.
y =sin(x)cos ( x
2
)
+ cos(x)sin
( x
2
)
and y =1 y = cos(x) −
√
3sin(x) andy =2
We repeat here the advice given when solving systems of nonlinear equations in section 8.7 –when
it comes to solving equations involving the trigonometric functions, it helps to just try something.
9 We are essentially ‘undoing’ the sum / difference formula for cosine or sine, depending on which form we use, so
this problem is actually closely related to the previous one!