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2. Solve $5\cos(2\theta) + 2 = 1$ over the interval $0 \le \theta < 2\pi$. Round intermediate steps to 4 decimal places, and round your final answers to 2 decimal places. List your solution set in ascending order.

          2. Solve $5\cos(2\theta) + 2 = 1$ over the interval $0 \le \theta < 2\pi$. Round intermediate steps to 4 decimal places, and round your final answers to 2 decimal places. List your solution set in ascending order.
        
2. Solve 5cos(2θ) + 2 = 1 over the interval 0 ≤θ < 2π. Round intermediate steps to 4 decimal places, and round your final answers to 2 decimal places. List your solution set in ascending order.

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Introductory and Intermediate Algebra for College Students 4th
Introductory and Intermediate Algebra for College Students 4th
Robert Blitzer 4th Edition
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Solve 5cos(20t)+2 = 1 over the interval 0 < 2t. Round intermediate steps to 4 decimal places, and round your final answers to 2 decimal places. List your solution set in ascending order.
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Transcript

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00:01 This problem wants us to use a calculator to solve the equation of the interval 0 less than or equal to theta less than 2 pi 4 5 plus secant of theta equal to 0.
00:10 And we are to use decimal notation and round our answers to three decimal places.
00:13 And what we will do to start is isolate our secant of theta by subtracting 5 to the right side.
00:19 So that's secant of theta equal to negative 5.
00:22 And what we need to realize now is that we're dealing with secant, which is the reciprocal of cosine.
00:26 So that's equivalent to 1 over cosine.
00:28 And cosine represents the x value.
00:31 And if we have 1 over an x value that's supposed to turn into negative 5, that means that our result needs to be something where x values are negative.
00:39 And x values are negative in the second and the third quadrant.
00:43 So what we're going to do to start is first find the solution in our first quadrant because that would be equivalent to our reference angle...
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