Solve the equation(GS general solution)
5sin^2x-3sinx-2=0
To solve the equation 5sin^2x - 3sinx - 2 = 0 for x, we can use the quadratic formula. The equation can be rewritten as follows:
5(sin^2x) - 3sinx - 2 = 0
Let's substitute sinx with a variable, say u:
5u^2 - 3u - 2 = 0
Using the quadratic formula, which states that for an equation ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac))/(2a)
For our equation, a = 5, b = -3, and c = -2. Let's plug them into the quadratic formula:
u = (-(-3) ± √((-3)^2 - 4 * 5 * (-2))) / (2 * 5)
u = (3 ± √(9 + 40)) / 10
u = (3 ± √49) / 10
u = (3 ± √7) / 10
So we have two possible values for u: u = (3 + √7) / 10 and u = (3 - √7) / 10.
Now, we substitute back sinx for u in the original equation:
sinx = (3 + √7) / 10 and sinx = (3 - √7) / 10
To find the general solution, we use the properties of the sine function. Since sinx is periodic with a period of 2π, we can add or subtract integer multiples of 2π from the solutions to obtain the general solution.
So, the general solution to the equation is:
x = arcsin((3 + √7) / 10) + 2πn and x = arcsin((3 - √7) / 10) + 2πn, where n is an integer.
To solve the equation 5sin^2x - 3sinx - 2 = 0, we will use the quadratic formula.
Step 1: Rearrange the equation in a standard quadratic form, where the equation can be written as: 5sin^2x - 3sinx - 2 = 0
Step 2: Replace sin^2x with (1 - cos^2x) using the identity sin^2x + cos^2x = 1, giving us:
5(1 - cos^2x) - 3sinx - 2 = 0
Step 3: Distribute the 5, giving us:
5 - 5cos^2x - 3sinx - 2 = 0
Step 4: Rearrange the equation and combine like terms:
-5cos^2x - 3sinx + 3 = 0
Step 5: Now, we have a quadratic equation in terms of cosx. Let's assume cosx as t to simplify the equation:
-5t^2 - 3sinx + 3 = 0
Step 6: Use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our equation, a = -5, b = -3sinx, and c = 3.
Substituting these values into the quadratic formula, we get:
t = (3sinx ± sqrt((-3sinx)^2 - 4 * (-5) * 3)) / (2 * (-5))
Simplifying further:
t = (3sinx ± sqrt(9sin^2x + 120)) / -10
Step 7: Solve for sinx:
Since sinx should be between -1 and 1, we ignore the negative solutions obtained from the quadratic formula. Let's solve for sinx when:
t = (3sinx + sqrt(9sin^2x + 120)) / -10
t = (3sinx + sqrt(9sin^2x + 120)) / -10 = 0
Step 8: Solve the equation using the information obtained:
3sinx + sqrt(9sin^2x + 120) = 0
Step 9: Rearrange the equation:
sqrt(9sin^2x + 120) = -3sinx
Step 10: Square both sides of the equation to eliminate the square root:
9sin^2x + 120 = 9sin^2x
Step 11: Subtract 9sin^2x from both sides of the equation:
120 = 0
Since 120 is not equal to 0, there are no real solutions to the original equation. Therefore, the equation 5sin^2x - 3sinx - 2 = 0 has no solutions in the real number system.
let y = sin x
5 y^2 -3 y - 2 = 0
(5 y +2)(y-1) = 0
y = 1 or y = -2/5
so
sin x = 1 ---> x = 90deg or 90+360 etc
sin x = -2/5 ---> x = -23.6 or 180+23.6 etc