Question

3 sin x - 2 cos^2 x +3=0

Answer 1

$3\sin x-2\cos^2x+3=0\\ 3\sin x-2(1-\sin^2x)+3=0\\ 3\sin x-2+2\sin^2x+3=0\\ 2\sin^2x+3\sin x+1=0$

Решим уравнение как квадратное уравнение относительно sin x:

$D=b^2-4ac=3^2-4\cdot 2\cdot 1=1$

$\sin x=\frac{-3+1}{2\cdot 2} =-0.5~~~\Rightarrow~~~ x=(-1)^{k+1}\cdot\frac{\pi}{6}+\pi k,k \in \mathbb{Z}\\ \\ \sin x=\frac{-3-1}{2\cdot 2}=-1~~~\Rightarrow~~~ x=-\frac{\pi}{2}+2\pi k,k \in \mathbb{Z}$

Answer 2

To solve the equation 3sin(x) - 2cos^2(x) + 3 = 0, we'll first make some trigonometric substitutions to simplify the equation. Recall the Pythagorean identity: sin^2(x) + cos^2(x) = 1. From this identity, we can rewrite cos^2(x) as 1 - sin^2(x).

Substitute 1 - sin^2(x) for cos^2(x) in the original equation:

3sin(x) - 2(1 - sin^2(x)) + 3 = 0

Now, let's simplify the equation further:

3sin(x) - 2 + 2sin^2(x) + 3 = 0

Combine the constants:

2sin^2(x) + 3sin(x) + 1 = 0

Now, this is a quadratic equation in terms of sin(x). We can solve it by factoring or using the quadratic formula.

Factor the quadratic equation:

(2sin(x) + 1)(sin(x) + 1) = 0

Now set each factor equal to zero and solve for sin(x):

1. 2sin(x) + 1 = 0 2sin(x) = -1 sin(x) = -1/2

2. sin(x) + 1 = 0 sin(x) = -1

So, we have two solutions for sin(x): sin(x) = -1/2 and sin(x) = -1.

Now, let's find the corresponding values of x:

1. sin(x) = -1/2 Using the unit circle or trigonometric ratios, we find two angles in the interval [0, 2π] where sin(x) = -1/2. Those angles are approximately x = 7π/6 and x = 11π/6.

2. sin(x) = -1 Again, using the unit circle or trigonometric ratios, we find x = 3π/2.

So, the solutions to the original equation 3sin(x) - 2cos^2(x) + 3 = 0 are: x = 7π/6, 11π/6, and 3π/2.