Question

Sin^2x+2sinx-3cos^2x+1=0

Answer 1

$sin^2 x+2sinx -3 cos^2 x+1=0$

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

$sin^2 x+2sinx-3+3sin^2 x+1=0$

$4sin^2 x+2sinx-2=0$

$2sin^2 x+sin x-1=0$

вводим замену $t=sin x$ , $-1 \leq t \leq 1$

$2t^2+t-1=0$

$D=1^2-4*2*(-1)=1+8=9=3^2$

$t_1=\frac{-1-3}{2*2}=-1$

$t_2=\frac{-1+3}{2*2}=\frac{1}{2}$

возвращаемся к замене

(1) $sin x=-1; x=-\frac{\pi}{2}+2*\pi*l$, l є Z

(2)$sin x=\frac{1}{2};$

$x=(-1)^k * arcsin \frac{1}{2} +\pi*k$, k є Z

$x=(-1)^k*\frac{\pi}{6}+\pi*k$, k є Z

ответ:$-\frac{\pi}{2}+2*\pi*l$, l є Z

$(-1)^k*\frac{\pi}{6}+\pi*k$, k є Z

Answer 2

To solve the trigonometric equation sin^2(x) + 2sin(x) - 3cos^2(x) + 1 = 0, we can use some trigonometric identities to simplify the equation and find its solutions. Here's how you can do it step by step:

Step 1: Use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to replace cos^2(x) in the equation.

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

Step 2: Distribute the -3 through the parentheses.

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

Step 3: Combine like terms.

4sin^2(x) + 2sin(x) - 2 = 0

Step 4: Divide the entire equation by 2 to simplify further.

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

Now we have a quadratic equation in terms of sin(x). Let's solve it by factoring:

Step 5: Factor the quadratic equation.

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

Now we have two factors that could equal zero:

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

Step 6: Solve each factor for x:

1. 2sin(x) - 1 = 0

Add 1 to both sides:

2sin(x) = 1

Divide by 2:

sin(x) = 1/2

To find the solutions for sin(x) = 1/2, we look for angles in the unit circle where the sine value is 1/2. These angles are π/6 and 5π/6, or in degrees approximately 30° and 150°.

2. sin(x) + 1 = 0

Subtract 1 from both sides:

sin(x) = -1

The solution for sin(x) = -1 is the angle -π/2 or -90°.

So, the solutions to the original trigonometric equation sin^2(x) + 2sin(x) - 3cos^2(x) + 1 = 0 are:

x = π/6 (30°) x = 5π/6 (150°) x = -π/2 (-90°)