trig question
2sin^2x+sinx-6=0
how to solve for x when the
restriction is x is greater than or equal to zero but less than 2pi
this is a quadratic, it might be easier to see if you let
sinx = t, then we have
2t^2 + t - 6 = 0 which factors to
(2t - 3)(t + 2) = 0
t = 3/2 or t = -2
so sinx = 3/2 or sinx = -2
both cases are not possible since sinx must be between -1 and +1
so your equation has no solution.
let y = sin x
the equation then becomes:
2y^2 + y - 6 = 0
then factor:
(2y - 3)(y + 2) = 0
y = 3/2 and y = -2
therefore:
sin x = 3/2, x = 2*pi*n + arcsin 3/2 ; and
sin x = -2 , x = 2*pi*n + arcsin (-2) , where n = integer
*but note that arcsin 3/2 and arcsin (-2) are undefined.
so there,, :)
To solve the trigonometric equation 2sin^2x + sinx - 6 = 0 with the given restriction, you can follow these steps:
Step 1: Rewrite the equation
Since the equation is quadratic in terms of sin(x), let's make a substitution to simplify it. Let's substitute sin(x) with another variable like t. So, the equation becomes:
2t^2 + t - 6 = 0
Step 2: Solve the quadratic equation
To solve this quadratic equation, you can either factorize it or use the quadratic formula. Factoring might not be easy in this case, so let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 2, b = 1, and c = -6.
Substituting these values into the quadratic formula, we get:
t = (-1 ± √(1^2 - 4(2)(-6))) / (2(2))
Simplifying further:
t = (-1 ± √(1 + 48)) / 4
t = (-1 ± √49) / 4
t1 = (-1 + 7) / 4 = 6 / 4 = 3 / 2
t2 = (-1 - 7) / 4 = -8 / 4 = -2
Step 3: Find the values of sin(x)
Since sin(x) = t, we have two possible values for sin(x):
sin(x) = 3 / 2
sin(x) = -2
However, sin(x) cannot be greater than 1 or less than -1, so discard the value sin(x) = -2.
Step 4: Find the angles in the given restriction
To find the values of x within the given restriction (x ≥ 0 and x < 2π), we need to find the angles whose sine is 3/2 and falls within this range.
The sine value of 3/2 corresponds to an angle outside the range of -1 to 1. However, there is a special case where sin(x) = 3/2, which occurs in the second quadrant where the sine function is positive.
In the second quadrant, sin(x) is positive, so we can find the angle using the inverse sine function:
x = arcsin(3/2)
Using a calculator, we find that arcsin(3/2) ≈ 1.047 radians. Note that this angle is less than 2π but greater than 0, satisfying our restriction.
Therefore, the solution to the equation 2sin^2x + sinx - 6 = 0 with the restriction x ≥ 0 and x < 2π is:
x = 1.047 radians.