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Answer:
18. x = {0, π/3, π, 5π/3, 2π}
19. x = {0, 2π}
Explanation:
You're supposed to use what you know about equation solving and trig functions to find the values of x that make these equations true. When the equation has a degree other than 1, you may need to use what you know about factoring and/or solving quadratic equations.
Inverse trig functions are helpful, but they don't always tell the whole story. You need to understand the behavior of each function over its whole period.
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18. This equation is easily factored.
-2sin(x)(1 -2cos(x)) = 0
The zero product rule tells you the product of these factors is zero only when one or more of the factors is zero. In other words, this resolves into the equations ...
Your knowledge of the sine function tells you the solutions to the first of these equations is x = 0, π, 2π. (in the range 0 ≤ x ≤ 2π)
The second equation can be rewritten as ...
1 = 2cos(x)
1/2 = cos(x)
Your knowledge of the cosine function tells you this is true for ...
x = π/3, 5π/3
So, all of the solutions to the given equation are ...
x = {0, π/3, π, 5π/3, 2π}
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19. Here, it is convenient to use a trig identity to make all of the variable terms be functions of the cosine.
sin(x)² = 1 - cos(x)² . . . . the trig identity we need
2 -(1 -cos(x)²) = 2cos(x) . . . . substitute for sin(x)²
1 + cos(x)² = 2cos(x) . . . . . . . simplify
cos(x)² -2cos(x) +1 = 0 . . . . . subtract 2cos(x), write as a quadratic in cos(x)
(cos(x) -1)² = 0 . . . . . . . . . . . factor (recognize the perfect square trinomial)
cos(x) = 1 . . . . . . . . . . . . . . take the square root, add 1
x = 0, 2π . . . . . . . . values of x for which this is true
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The attachments show the solutions found using a graphing calculator. When solving these by graphing, it is generally most convenient to rewrite the equation to the form f(x) = 0. This can be done by subtracting the right-side expression, for example, as we did in the second attachment. That way, the solutions are the x-intercepts, which most graphing calculators can find easily.