Asked by Tiana
Find the general solution for x if cos2x + sin3x = sinx
Answers
Answered by
Reiny
Nice question !
cos2x + sin3x = sinx
let's change everything to sinx
(1 - 2sin^2 x) + (3sinx - 4sin^3 x) - sinx = 0
4sin^3 x + 2sin^2 x - 2sinx -1 = 0
2sin^2 x (2sinx + 1) - 1(2sinx + 1) = 0
(2sinx + 1) (2sin^2x -1) = 0
sinx = -1/2 or sin^2 x = 1/2
for sinx = -1/2
x = <b>7π/6 or 11π/6</b> --------- ( 210° or 330°)
for sin^2 x = 1/2
sinx = ± 1/√2
x = <b>π/4 , 3π/4 , 5π/4 , 7π/4</b> ------- ( 45°, 135°, 225° , 315°)
for general solutions take each of the above answers and add 2kπ
e.g. 7π/6 + 2kπ , where k is an integer.
cos2x + sin3x = sinx
let's change everything to sinx
(1 - 2sin^2 x) + (3sinx - 4sin^3 x) - sinx = 0
4sin^3 x + 2sin^2 x - 2sinx -1 = 0
2sin^2 x (2sinx + 1) - 1(2sinx + 1) = 0
(2sinx + 1) (2sin^2x -1) = 0
sinx = -1/2 or sin^2 x = 1/2
for sinx = -1/2
x = <b>7π/6 or 11π/6</b> --------- ( 210° or 330°)
for sin^2 x = 1/2
sinx = ± 1/√2
x = <b>π/4 , 3π/4 , 5π/4 , 7π/4</b> ------- ( 45°, 135°, 225° , 315°)
for general solutions take each of the above answers and add 2kπ
e.g. 7π/6 + 2kπ , where k is an integer.
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