Question
Asked Nov 20, 2019
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Solve the trigonometric equation in the interval [0, 27t). Give the exact value, if possible; otherwise, round your answer to two decimal places. (Enter your answers as a comma-separated list.)
2 cos2(e)19 cos(e) 9 = 0
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Solve the trigonometric equation in the interval [0, 27t). Give the exact value, if possible; otherwise, round your answer to two decimal places. (Enter your answers as a comma-separated list.) 2 cos2(e)19 cos(e) 9 = 0

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Expert Answer

Step 1

The given trigonometric equation is shown below.

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2 cos2 e19 cose+9 = 0

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Step 2

The above equation is quadratic in cos x.

Now substitute cos θ = u and solve the quadratic equation using quadratic formula.

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2u19u9=0 Quadratic formula: The solution of a quadratic equation ax bx +x 0 is: -btb-4ac 2a Therefore -19t v19-4.2.9 2.2 -19+289 4 -19 17 = 4 1 -9 1 -9 2 cose But cose can never be less than - 1, thus ignoring - 9 1 .coSe 2

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Step 3

Now solving...

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1 cos x 2 1 cos.x cos cos 3 3 2 27T cos x cos- 3 cos (-x)cos.x If cos x cos y 27T then x 2n7t y 3 2T 4T 107T It gives, x 3 3 3

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