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Solve for x in the following equations.

0 votes
1) 2sinx-1=0

2)2sin^2-sinx-1=0
asked Jun 24, 2013 in TRIGONOMETRY by dkinz Apprentice

2 Answers

0 votes

1) 2sinx-1=0

Add 1 to each side.
2sinx = 1

Divide each side by 2.
sinx = 1/2

sinx = sin30º

x = 30º

The solution is x = π/6

answered Jun 24, 2013 by anonymous
0 votes

1ans)   2sinx - 1 = 0

             2sinx = 1

             sinx = 1/2

             x = sin-1(1/2)

             x = sin-1(sinπ/6)

             x = π/6

2ans)   2sin^2x  -  sinx - 1 = 0

            2sin^2x  -  2sinx + sinx  - 1 = 0

            2sinx(sinx - 1)  + 1(sinx - 1) =0

           Take common factor (sinx - 1) on each side

             (sinx - 1) (2sinx + 1) = 0

             sinx - 1 = 0                  2sinx + 1 = 0

             sinx = 1                        2sinx + 1 = 0

              x = sin-1( 1)                sinx = -1/2

              x = sin-1(1)                  x = sin-1(-1/2)

              x = sin-1(sinπ/2)         x = sin-1(-sinπ/3)

              x = π/2                         x = -π/6

       The solution set for x is {π/6 , π/2 , -π/6}

answered Jun 24, 2013 by jouis Apprentice

(1).

sin(x) = 1/2

sin(x) = sin(π/6)

General solution : If sin(θ) = sin(α), then θ = nπ + (-1)nα, where n is an integer.

If α = π/6 then x = nπ + (-1)n(π/6).

The general solution is x = nπ + (-1)n(π/6).

 

(2).

(sinx - 1) (2sinx + 1) = 0

sin(x) = 1 and sin(x) = -1/2

sin(x) = sin(π/2) and sin(x) = sin(-π/6)

General solution : If sin(θ) = sin(α), then θ = nπ + (-1)nα, where n is an integer.

If α = π/2 then x = nπ + (-1)n(π/2).

If α = -π/6 then x = nπ + (-1)n(-π/6) = nπ - (-1)n(π/6).

The general solutions are x = nπ + (-1)n(π/2) and x = nπ - (-1)n(π/6).

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