744 CHAPTER 7 Trigonometric Identities and Equations Equations with Half-Angles EXAMPLE 6 Solving an Equation with a Half-Angle Solve 2 sin x 2 = 1 (a) over the interval 30, 2p2 (b) for all solutions. SOLUTION (a) To solve over the interval 30, 2p2, we must have 0 … x 62p. The corresponding inequality for x 2 is 0 … x 2 6p. Divide each part by 2. To find all values of x 2 over the interval 30, p2 that satisfy the given equation, first solve for sin x 2 . 2 sin x 2 = 1 Original equation sin x 2 = 1 2 Divide by 2. The two numbers over the interval 30, p2 with sine value 1 2 are p 6 and 5p 6 . x 2 = p 6 or x 2 = 5p 6 Definition of inverse sine x = p 3 or x = 5p 3 Multiply by 2. The solution set over the given interval is Ep 3 , 5p 3 F . (b) The argument x 2 in the expression sin x 2 can also be written 1 2 x to see that the value of b in sin bx is 1 2 . From earlier work we know that the period is 2p b , so we replace b with 1 2 in this expression and perform the calculation. Here the period is 2p 1 2 = 2p , 1 2 = 2p # 2 = 4p. All solutions are found by adding integer multiples of 4p. E p 3 + 4np, 5p 3 + 4np, where n is any integerF S Now Try Exercises 77 and 91. −2 2 0 2p The x-intercepts correspond to the solutions found in Example 6(a). Using Xscl = p 3 makes it possible to support the exact solutions by counting the tick marks from 0 on the graph. CAUTION Because 2 is not a factor of cos 2x, cos 2x 2 ≠cos x. In Example 7, on the next page, we changed cos 2x to a function of x alone using an identity.
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