solve: (square root of 3) sin(theta) + cos(theta) = 1
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To solve this equation, we'll rearrange it and apply some trigonometric identities.
First, let's move the cosine term to the other side of the equation:
(square root of 3) sin(theta) = 1 - cos(theta)
Next, let's square both sides of the equation to remove the square root:
3*sin^2(theta) = (1 - cos(theta))^2
Expanding the right side using the binomial formula:
3*sin^2(theta) = 1 - 2*cos(theta) + cos^2(theta)
Next, let's use the trigonometric identity sin^2(theta) + cos^2(theta) = 1:
3*(1 - cos^2(theta)) = 1 - 2*cos(theta) + cos^2(theta)
3 - 3*cos^2(theta) = 1 - 2*cos(theta) + cos^2(theta)
Rearranging terms:
4*cos^2(theta) - 2*cos(theta) + 2 = 0
Now, let's substitute x = cos(theta) to get a quadratic equation:
4*x^2 - 2*x + 2 = 0
Using the quadratic formula:
x = (-b ± sqrt(b^2 - 4*a*c)) / (2*a)
where a = 4, b = -2, and c = 2.
Plugging in the values:
x = (-(-2) ± sqrt((-2)^2 - 4*4*2)) / (2*4)
x = (2 ± sqrt(4 - 32)) / 8
x = (2 ± sqrt(-28)) / 8
Since the discriminant (b^2 - 4*a*c) is negative, there are no real solutions for x, which means there are no solutions for cos(theta).
Therefore, the original equation (square root of 3) sin(theta) + cos(theta) = 1 has no solutions.