6 in examining the equation sin2 x 3 cos x not only

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6. In examining the equation sin(2 x ) = 3 cos( x ), not only do we have different circular func- tions involved, namely sine and cosine, we also have different arguments to contend with, namely 2 x and x . Using the identity sin(2 x ) = 2 sin( x ) cos( x ) makes all of the arguments the same and we proceed as we would solving any nonlinear equation – gather all of the nonzero terms on one side of the equation and factor. sin(2 x ) = 3 cos( x ) 2 sin( x ) cos( x ) = 3 cos( x ) (Since sin(2 x ) = 2 sin( x ) cos( x ).) 2 sin( x ) cos( x ) - 3 cos( x ) = 0 cos( x )(2 sin( x ) - 3) = 0 from which we get cos( x ) = 0 or sin( x ) = 3 2 . From cos( x ) = 0, we obtain x = π 2 + πk for integers k . From sin( x ) = 3 2 , we get x = π 3 +2 πk or x = 2 π 3 +2 πk for integers k . The answers 7 As always, experience is the greatest teacher here! 8 As always, when in doubt, write it out!
10.7 Trigonometric Equations and Inequalities 865 which lie in [0 , 2 π ) are x = π 2 , 3 π 2 , π 3 and 2 π 3 . We graph y = sin(2 x ) and y = 3 cos( x ) and, after some careful zooming, verify our answers. y = cos(3 x ) and y = cos(5 x ) y = sin(2 x ) and y = 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. If we stare at 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 + 4 π 3 k for integers k . Two of these solutions lie in [0 , 2 π ): x = π 3 and x = 5 π 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 ) - 3 sin( 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 = 5 π 6 . Hence, we can rewrite the equation cos( x ) - 3 sin( x ) = 2 as 2 sin ( x + 5 π 6 ) = 2, or sin ( x + 5 π 6 ) = 1. Solving the latter, we get x = - π 3 + 2 πk for integers k . Only one of these solutions, x = 5 π 3 , which corresponds to k = 1, lies in [0 , 2 π ). Geometrically, we see that y = cos( x ) - 3 sin( x ) and y = 2 intersect just once, supporting our answer. y = sin( x ) cos ( x 2 ) + cos( x ) sin ( x 2 ) and y = 1 y = cos( x ) - 3 sin( x ) and y = 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!
866 Foundations of Trigonometry Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we’ve used in the past to solve the inequalities.

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