How do you solve for x when 2sin^2(x-(pi/6)=1?

Question

How do you solve for x when 2sin^2(x-(pi/6)=1?

Answer 1

sin^2(x-pi/6) = 1/2

sin(x-pi/6) = sqrt(1/2)

x - pi/6 = sin-1(sqrt(1/2)) = pi/4

x = pi/4 + pi/6

Answer 2

Solve for x.

2sin²(x - π/6) = 1

sin²(x - π/6) = 1/2

sin(x - π/6) = ±1/√2

x - π/6 = π/4, 3π/4, 5π/4, 7π/4

x = 5π/12, 11π/12, 17π/12, 23π/12

On the interval [0, 2π).

Answer 3

2sin²(x-π/6) = 1

sin²(x-π/6) = 1/2

sin(x-π/6) = ± √(1/2) = ± √2/2

(x-π/6) = asin(± √2/2)

(x-π/6) = ± π/4 and ± 3π/4

x = π/4 + π/6 = 5π/12  and -π/4 + π/6 = -π/12

x = 3π/4 + π/6 = 11π/12 and -3π/4 + π/6 = -7π/12

Answers: 5π/12,  11π/12, and -π/12, -7π/12

Answer 4

2sin^2[x-(pi/6)] = 1

=> sin^2[x-(pi/6)] = 1/2 = sin^2[pi/4] ( since sin(pi/4) = 1/root2 )

=> x - (pi/6) = pi/4

=> x = 5pi/12

there u go.. :)