Solve the equation on the interval [0,2pi).
2sin^2x-3sinx+1=0
(2sinx+1)(sinx+1)
I don't think I did the factoring correctly. When I multiply it out to double check I get 2sin^2x+3sinx+1
Notice that the original had a negative term.
Your factored form is all positive, so you know for sure it is not correct.
how about
(2sinx - 1)(sinx - 1) = 0 ? , yes that works
so sinx = 1/2 or sinx = 1
let's do the easy one, sinx x = 1 , so x = π/2
for sinx = 1/2, we know that sine is positive in quads I and II by the CAST rule
so x = π/6 or x = π-π/6 = 5π/6
so x = π/2, π/6, 5π/6
you just got the signs reversed
(2 sinx - 1)(sinx - 1)
To solve the equation 2sin^2x - 3sinx + 1 = 0, let's try factoring it correctly:
The equation is in the form of a quadratic equation, so we can use factoring.
The factors of 2sin^2x are (2sinx)(sinx) or (sinx)(2sinx).
The factors of 1 are (1)(1).
To get -3sinx as the middle term, we need to find two numbers whose product is (2sinx)(1) = 2sinx and whose sum is -3sinx.
Since -2sinx and -1sinx multiply to give 2sinx and add up to -3sinx, we can rewrite the equation as follows:
2sin^2x - 2sinx - sinx + 1 = 0
Now, let's group the terms:
(2sin^2x - 2sinx) + (-sinx + 1) = 0
Factor out common terms from each group:
2sinx(sinx - 1) - 1(sinx - 1) = 0
Now, observe that we have a common binomial factor (sinx - 1):
(sinx - 1)(2sinx - 1) = 0
Now, set each factor to zero and solve for x:
sinx - 1 = 0 or 2sinx - 1 = 0
If sinx - 1 = 0, adding 1 to each side gives:
sinx = 1
On the interval [0,2π), the solution for sinx = 1 is x = π/2.
If 2sinx - 1 = 0, adding 1 to each side gives:
2sinx = 1
Divide by 2:
sinx = 1/2
On the interval [0,2π), the solutions for sinx = 1/2 are x = π/6 and x = 5π/6.
So, the solutions for the equation 2sin^2x - 3sinx + 1 = 0 on the interval [0,2π) are x = π/2, x = π/6, and x = 5π/6.
To solve the equation 2sin^2(x) - 3sin(x) + 1 = 0, you can indeed use factoring to find the solutions. However, it seems that there might be a mistake in your factoring. Let's go through the correct factoring process.
To begin, notice that the equation is a quadratic equation in terms of sin(x). Let's rewrite it for clarity:
2(sin^2(x)) - 3sin(x) + 1 = 0
To factor this quadratic equation, we need to find two binomials whose product is equal to the equation. In this case, the quadratic equation takes the form:
(ax + b)(cx + d) = 0
To solve for a, b, c, and d, we need to find values that satisfy the following conditions:
ac = 2
ad + bc = -3
bd = 1
In this particular equation, it's not obvious to find two values of ac and bd that multiply to 2 and 1, respectively, and satisfy the middle term. Hence, factoring might not be the best approach to solve this equation. Instead, we can try using the quadratic formula.
The quadratic formula states that for any quadratic equation in the form ax^2 + bx + c = 0, the solutions x can be found using:
x = (-b ± √(b^2 - 4ac)) / (2a)
Applying this formula to the equation 2sin^2(x) - 3sin(x) + 1 = 0, we can see that a = 2, b = -3, and c = 1. Thus, the possible solutions for sin(x) are:
sin(x) = (-(-3) ± √((-3)^2 - 4*2*1)) / (2*2)
simplifying further:
sin(x) = (3 ± √(9-8)) / 4
sin(x) = (3 ± √1) / 4
Now, we can evaluate the solutions:
For the positive case:
sin(x) = (3 + 1) / 4
sin(x) = 4 / 4
sin(x) = 1
For the negative case:
sin(x) = (3 - 1) / 4
sin(x) = 2 / 4
sin(x) = 1/2
Hence, we found two potential solutions for sin(x).
However, it's worth noting that we were solving for sin(x), not x itself. Therefore, to find the values of x that correspond to these solutions, you need to use the inverse sine function, sin^(-1), or arcsin.
Using arcsin:
x = arcsin(1)
x = pi / 2
x = arcsin(1/2)
x = pi / 6
Therefore, the solutions to the equation 2sin^2(x) - 3sin(x) + 1 = 0 on the interval [0, 2pi) are x = pi/2 and x = pi/6.