To solve this equation, we can use the trigonometric identity that relates cosine and sine:
cos^2(x) - sin^2(x) = 1
Substitute cos^2(x) with 1 - sin^2(x):
1 - sin^2(x) - 2sin(x) = -(1/4)
Rearrange the terms and get everything on one side:
sin^2(x) + 2sin(x) - 1/4 = 0
Now, we can solve this quadratic equation for sin(x):
Let's substitute sin(x) with y:
y^2 + 2y - 1/4 = 0
Now, solve the quadratic equation for y:
y = (-b ± sqrt(b^2 - 4ac))/(2a)
y = (-2 ± sqrt(2^2 - 41(-1/4)))/(2*1)y = (-2 ± sqrt(4 + 1))/2y = (-2 ± sqrt(5))/2
Now, substitute back sin(x) to y:
sin(x) = (-2 ± sqrt(5))/2
Therefore, the solutions for the equation are:
sin(x) = (-2 + √5)/2sin(x) = (-2 - √5)/2
To solve this equation, we can use the trigonometric identity that relates cosine and sine:
cos^2(x) - sin^2(x) = 1
Substitute cos^2(x) with 1 - sin^2(x):
1 - sin^2(x) - 2sin(x) = -(1/4)
Rearrange the terms and get everything on one side:
sin^2(x) + 2sin(x) - 1/4 = 0
Now, we can solve this quadratic equation for sin(x):
Let's substitute sin(x) with y:
y^2 + 2y - 1/4 = 0
Now, solve the quadratic equation for y:
y = (-b ± sqrt(b^2 - 4ac))/(2a)
y = (-2 ± sqrt(2^2 - 41(-1/4)))/(2*1)
y = (-2 ± sqrt(4 + 1))/2
y = (-2 ± sqrt(5))/2
Now, substitute back sin(x) to y:
sin(x) = (-2 ± sqrt(5))/2
Therefore, the solutions for the equation are:
sin(x) = (-2 + √5)/2
sin(x) = (-2 - √5)/2