Answer :
To solve the equation [tex]\(2 \sin^2(\theta) + \sin(\theta) - 1 = 0\)[/tex], we can follow these steps:
1. Let [tex]\(u = \sin(\theta)\)[/tex]:
This transforms the given trigonometric equation into a standard quadratic equation:
[tex]\[ 2u^2 + u - 1 = 0 \][/tex]
2. Solve the quadratic equation [tex]\(2u^2 + u - 1 = 0\)[/tex]:
For a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], we use the quadratic formula:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -1\)[/tex].
Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9 \][/tex]
Find the square root of the discriminant:
[tex]\[ \sqrt{9} = 3 \][/tex]
Substitute back into the quadratic formula to find the values of [tex]\(u\)[/tex]:
[tex]\[ u_1 = \frac{-1 + 3}{2 \cdot 2} = \frac{2}{4} = 0.5 \][/tex]
[tex]\[ u_2 = \frac{-1 - 3}{2 \cdot 2} = \frac{-4}{4} = -1 \][/tex]
3. Find the corresponding values of [tex]\(\theta\)[/tex] for each [tex]\(u\)[/tex]:
We now consider [tex]\(\sin(\theta) = u\)[/tex].
For [tex]\(u = 0.5\)[/tex]:
[tex]\[ \theta_{1,1} = \arcsin(0.5) = 0.5236 \quad \text{(in radians)} \][/tex]
Since [tex]\(\sin(\theta) = \sin(\pi - \theta)\)[/tex], the second solution in this range for [tex]\(\sin(\theta) = 0.5\)[/tex] is:
[tex]\[ \theta_{1,2} = \pi - 0.5236 = 2.6180 \quad \text{(in radians)} \][/tex]
For [tex]\(u = -1\)[/tex]:
[tex]\[ \theta_{2,1} = \arcsin(-1) = -1.5708 \quad \text{(in radians)} \][/tex]
Since [tex]\(\sin(\theta) = \sin(\pi - \theta)\)[/tex], the second solution in this range for [tex]\(\sin(\theta) = -1\)[/tex] is:
[tex]\[ \theta_{2,2} = \pi - (-1.5708) = 4.7124 \quad \text{(in radians)} \][/tex]
Therefore, the solutions to the equation [tex]\(2 \sin^2(\theta) + \sin(\theta) - 1 = 0\)[/tex] are:
[tex]\[ u_1 = 0.5, \quad u_2 = -1.0 \][/tex]
[tex]\[ \theta_{1,1} = 0.5236, \quad \theta_{1,2} = 2.6180 \][/tex]
[tex]\[ \theta_{2,1} = -1.5708, \quad \theta_{2,2} = 4.7124 \][/tex]
These values are the results in radians.
1. Let [tex]\(u = \sin(\theta)\)[/tex]:
This transforms the given trigonometric equation into a standard quadratic equation:
[tex]\[ 2u^2 + u - 1 = 0 \][/tex]
2. Solve the quadratic equation [tex]\(2u^2 + u - 1 = 0\)[/tex]:
For a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], we use the quadratic formula:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -1\)[/tex].
Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9 \][/tex]
Find the square root of the discriminant:
[tex]\[ \sqrt{9} = 3 \][/tex]
Substitute back into the quadratic formula to find the values of [tex]\(u\)[/tex]:
[tex]\[ u_1 = \frac{-1 + 3}{2 \cdot 2} = \frac{2}{4} = 0.5 \][/tex]
[tex]\[ u_2 = \frac{-1 - 3}{2 \cdot 2} = \frac{-4}{4} = -1 \][/tex]
3. Find the corresponding values of [tex]\(\theta\)[/tex] for each [tex]\(u\)[/tex]:
We now consider [tex]\(\sin(\theta) = u\)[/tex].
For [tex]\(u = 0.5\)[/tex]:
[tex]\[ \theta_{1,1} = \arcsin(0.5) = 0.5236 \quad \text{(in radians)} \][/tex]
Since [tex]\(\sin(\theta) = \sin(\pi - \theta)\)[/tex], the second solution in this range for [tex]\(\sin(\theta) = 0.5\)[/tex] is:
[tex]\[ \theta_{1,2} = \pi - 0.5236 = 2.6180 \quad \text{(in radians)} \][/tex]
For [tex]\(u = -1\)[/tex]:
[tex]\[ \theta_{2,1} = \arcsin(-1) = -1.5708 \quad \text{(in radians)} \][/tex]
Since [tex]\(\sin(\theta) = \sin(\pi - \theta)\)[/tex], the second solution in this range for [tex]\(\sin(\theta) = -1\)[/tex] is:
[tex]\[ \theta_{2,2} = \pi - (-1.5708) = 4.7124 \quad \text{(in radians)} \][/tex]
Therefore, the solutions to the equation [tex]\(2 \sin^2(\theta) + \sin(\theta) - 1 = 0\)[/tex] are:
[tex]\[ u_1 = 0.5, \quad u_2 = -1.0 \][/tex]
[tex]\[ \theta_{1,1} = 0.5236, \quad \theta_{1,2} = 2.6180 \][/tex]
[tex]\[ \theta_{2,1} = -1.5708, \quad \theta_{2,2} = 4.7124 \][/tex]
These values are the results in radians.