Question

2sin^2x + 3sin x + 1 = 0.​

Answer 1

2sin²x+3sinx+ 1 = 0.

-1≤sinx≤1

Решая квадратное уравнение относительно синуса по Виету, получим sinx=-1; sinx=-1/2

Если  sinx=-1, то х=-π/2+2πк; к∈Z

Если же sinx=-1/2, то х=(-1)ⁿarcsin(-1/2)+πn n∈Z;  х=(-1)ⁿ(-π/6)+πn n∈Z;

х=(-1)ⁿ ⁺¹(π/6)+πn n∈Z;

Answer 2

Решение:

$2\sin ^2(x)+3\sin (x)+1=0\\\big | \: \sin (x)=t\\2t^2+3t+1=0\\2t^2+2t+t+1=0\\2t(t+1)+1(t+1)=0\\(t+1)(2t+1)=0\\t_1=-1; \quad t_2=-\frac{1}{2}\\\big |\: t=\sin (x)\\\begin{bmatrix}\sin (x) = -1 \\ \sin (x)=-\frac{1}{2} \end.$

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Градусы: $x=210^{\circ}+360^{\circ}n,\:x=330^{\circ}+360^{\circ}n,\:x=270^{\circ}+360^{\circ}n$

Answer 3

The given equation is a quadratic equation in terms of the variable sin(x). To solve for sin(x), you can use the quadratic formula:

For a quadratic equation of the form ax^2 + bx + c = 0, the solutions are given by: x = (-b ± √(b^2 - 4ac)) / (2a)

In your equation, substitute sin(x) for x, a = 2, b = 3, and c = 1:

sin(x) = (-3 ± √(3^2 - 4 * 2 * 1)) / (2 * 2) sin(x) = (-3 ± √(9 - 8)) / 4 sin(x) = (-3 ± √1) / 4

This simplifies to two possible solutions for sin(x):

1. sin(x) = (-3 + 1) / 4 = -2 / 4 = -0.5
2. sin(x) = (-3 - 1) / 4 = -4 / 4 = -1

However, the sine function's range is between -1 and 1, so the second solution, sin(x) = -1, is not valid.

Thus, the only valid solution is sin(x) = -0.5. To find the corresponding angle(s), you can use the inverse sine function (also known as arcsin or sin^-1). In degrees, you would have:

x = arcsin(-0.5) x ≈ -30° or 210°

So, the solutions for the equation 2sin^2x + 3sin x + 1 = 0 are x ≈ -30° and x ≈ 210°, considering the periodic nature of the sine function.