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Ksenya-84 [330]
4 years ago
8

2sin^2x-sinx-1=0, solve for x

Mathematics
2 answers:
Alla [95]4 years ago
6 0
2 sin² x - sin x - 1 = 0
u = sin x ( substitution )
2 u² - u - 1=0
2 u² - 2 u + u - 1 = 0
2 u ( u - 1 ) + ( u - 1 ) = 0
( u - 1 ) + ( 2 u + 1 ) = 0
( sin x - 1 ) ( 2 sin x + 1 ) = 0
sin x - 1 = 0                 or:   2 sin x + 1 = 0
sin x = 1                              sin x = -1/2
x 1 = π/2 + 2 kπ                x 2 = 7π/6 + 2 kπ,  x 3 = - π/6 + 2 kπ, k ∈ Z
k ∈ Z 
Gemiola [76]4 years ago
6 0
                                     2sin²(x) - sin(x) - 1 = 0
                      2sin²(x) - 2sin(x) + sin(x) - 1 = 0
2sin(x)[sin(x)] - 2sin(x)[1] + 1[sin(x)] - 1[1] = 0
                 2sin(x)[sin(x) - 1] + 1[sin(x) - 1] = 0
                               [2sin(x) + 1][sin(x) - 1] = 0
                  2sin(x) + 1 = 0    or    sin(x) - 1 = 0
                               - 1  - 1                    + 1 + 1
                        2sin(x) = -1               sin(x) = 1
                            2          2       sin⁻¹[sin(x)] = sin⁻¹(1)
                          sin(x) = -0.5                   x = 90, 450
                  sin⁻¹[sin(x)] = sin⁻¹(-0.5)
                                 x = -30, 210
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