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Solve for the exact solutions in the interval

List your answers separated by a comma, if it has no real solutions, enter DNE.

Solve for the exact solutions in the interval List your answers separated by a comma-example-1
User Jon Lee
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1 Answer

4 votes

Answer:


√(2) \sin(3x) - 1 = 0


√(2) \sin(3x) = 1


\sin(3x) = ( √(2) )/(2)

3x = π/4 + 2kπ or 3x = 3π/4 + 2kπ

x = π/12 + 2kπ/3 or x = π/4 + 2kπ/3

0 < π/12 + 2kπ/3 <

0 < 1/12 + 2k/3 < 2

0 < 1 + 8k < 24

-1 < 8k < 23, so k = 0, 1, 2

x = π/12, 3π/4, 17π/12

0 < π/4 + 2kπ/3 <

0 < 1/4 + 2k/3 < 2

0 < 3 + 8k < 24

-3 < 8k < 21, so k = 0, 1, 2

x = π/4, 11π/12, 19π/12

User Belindanju
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