Math  /  Trigonometry

QuestionSolve 2sin2x=1.6,0x2π2 \sin 2 x=1.6, \quad 0 \leq x \leq 2 \pi

Studdy Solution
Determine the values of kk such that 0x2π0 \leq x \leq 2\pi:
For x=0.46365+kπx = 0.46365 + k\pi:
- k=0k = 0: x=0.46365x = 0.46365 - k=1k = 1: x=0.46365+π3.60524x = 0.46365 + \pi \approx 3.60524
For x=1.10715+kπx = 1.10715 + k\pi:
- k=0k = 0: x=1.10715x = 1.10715 - k=1k = 1: x=1.10715+π4.24874x = 1.10715 + \pi \approx 4.24874
The solutions for xx within the interval 0x2π0 \leq x \leq 2\pi are:
x0.46365,1.10715,3.60524,4.24874 x \approx 0.46365, \, 1.10715, \, 3.60524, \, 4.24874

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