solve
3sin^2(x)=5sin(x)+2 (0≤x<π)
3sin^2(x)=5sin(x)+2
for readability, let u = sin(x)
3u^2 - 5u - 2 = 0
(u-2)(3u+1) = 0
so, we need
sin(x) = 2 -- ain't likely, or
sin(x) = -1/3
now, sin(x) >= 0 for 0≤x<π, so there is no solution
3sin^2 x - 5sin x - 2 = 0
(3sinx + 1)(sinx - 2) = 0
sinx = -1/3 or sinx = 2, the last part is not possible
if sinx = -1/3
x must be in quadrants III or IV , but 0≤x ≤ π
so there is no solution in your given domain
To solve the equation 3sin^2(x) = 5sin(x) + 2, where 0 ≤ x < π, we can use a substitution. Let's introduce a new variable, say u, which is equal to sin(x):
u = sin(x)
Now, we can rewrite the equation in terms of u:
3u^2 = 5u + 2
This is now a quadratic equation in terms of u. Rearranging terms, we have:
3u^2 - 5u - 2 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula here to find the values of u:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Applying this to our equation, with a = 3, b = -5, and c = -2, we find:
u = (-(-5) ± √((-5)^2 - 4(3)(-2))) / (2(3))
Simplifying further:
u = (5 ± √(25 + 24)) / 6
u = (5 ± √49) / 6
u = (5 ± 7) / 6
This gives us two possible solutions for u:
u₁ = (5 + 7) / 6 = 2
u₂ = (5 - 7) / 6 = -1/3
Since we defined u as sin(x), we need to find the values of x that correspond to these solutions for u.
For u = 2:
sin(x) = 2
There are no values of x that satisfy this equation, as sin(x) is always between -1 and 1.
For u = -1/3:
sin(x) = -1/3
To find the values of x, we can take the inverse sine (arcsin) of both sides:
x = arcsin(-1/3)
Using a calculator or a table of trigonometric functions, we can find the value of arcsin(-1/3) to be approximately -0.3398.
Therefore, the solution to the equation 3sin^2(x) = 5sin(x) + 2 in the given interval is:
x ≈ -0.3398 (0 ≤ x < π)