Question

18sin^2x +3 sinx-10=0

Answer 1

Пусть sinx=f, тогда
18F^2+3f-10=0
д=9-4×18×(-10)=9+720=729
f1=-3+27/36= 24/36= 2/3
f2=-3-27/36=-30/36=-5/6
sinf=2/3
f1=(-1)^n×arcsin2/3+пn, n принадлежит Z
sinf=-5/6
f=(-1)^n×(п-arcsin 5/6)+пn, n принадлежит Z

Answer 2

To solve the quadratic equation 18sin²x + 3sinx - 10 = 0, we can use a substitution. Let's denote sin(x) as a variable, say u. Then the equation becomes:

18u² + 3u - 10 = 0

Now, we can solve this quadratic equation for u using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula, which states that for an equation of the form ax² + bx + c = 0, the solutions for x can be found using:

x = (-b ± √(b² - 4ac)) / (2a)

For our equation 18u² + 3u - 10 = 0, we have a = 18, b = 3, and c = -10. Plugging these values into the quadratic formula, we get:

u = (-3 ± √(3² - 4 * 18 * -10)) / (2 * 18)

Simplifying further:

u = (-3 ± √(9 + 720)) / 36 u = (-3 ± √729) / 36 u = (-3 ± 27) / 36

Now we have two possible solutions for u:

1. u = (-3 + 27) / 36 = 24 / 36 = 2/3
2. u = (-3 - 27) / 36 = -30 / 36 = -5/6

Since sin(x) is limited to the range [-1, 1], the second solution u = -5/6 is not valid. Therefore, sin(x) = 2/3.

To find the values of x, we can use the inverse sine function (also known as arcsin or sin^(-1)). Taking the inverse sine of both sides, we have:

x = arcsin(2/3)

The inverse sine function returns an angle whose sine is the given value (2/3 in this case). Depending on the range of angles you're interested in, you may need to consider multiple solutions. In the interval [-π/2, π/2], the principal value of arcsin(2/3) is approximately 0.7297 radians or approximately 41.81 degrees.

Therefore, one solution to the equation 18sin²x + 3sinx - 10 = 0 is:

x ≈ 0.7297 radians (or ≈ 41.81 degrees)

Please note that there may be other solutions outside the principal value range, considering the periodic nature of the sine function.