Question

Решить: 2sin^2x - 2cosx = 5/2

Answer 1

Ответ:

$x=\pi \pm \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}$

Объяснение:

$\begin{matrix}2sin^2(x)-2cos(x)=\frac{5}{2}\\2-2cos^2(x)-2cos(x)=\frac{5}{2}\\2cos^2(x)+2cos(x)+\frac{1}{2}=0\\4cos^2(x)+4cos(x)+1=0\\(2cos(x)+1)^2=0\\cos(x)=-\frac{1}{2}\\x=\pi \pm \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}\end{matrix}$

Answer 2

To solve the trigonometric equation 2sin^2(x) - 2cos(x) = 5/2, we can use the trigonometric identities to simplify the equation.

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

Step 2: Express sin^2(x) in terms of cos(x) using the Pythagorean identity: sin^2(x) = 1 - cos^2(x).

Now, substitute sin^2(x) in the original equation:

2(1 - cos^2(x)) - 2cos(x) = 5/2

Step 3: Distribute and rearrange:

2 - 2cos^2(x) - 2cos(x) = 5/2

Step 4: Move everything to one side to form a quadratic equation:

2cos^2(x) + 2cos(x) - 1/2 = 0

Step 5: To solve the quadratic equation, let's make a substitution: Let y = cos(x).

Now the equation becomes:

2y^2 + 2y - 1/2 = 0

Step 6: Solve the quadratic equation. We can use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / 2a

where a = 2, b = 2, and c = -1/2.

y = (-(2) ± √(2^2 - 4 * 2 * (-1/2))) / 2 * 2

y = (-2 ± √(4 + 4)) / 4

y = (-2 ± √(8)) / 4

Step 7: Simplify the expression for y:

y = (-2 ± 2√2) / 4

Step 8: Reduce the fraction:

y = (-1 ± √2) / 2

Now, remember that y = cos(x):

cos(x) = (-1 ± √2) / 2

Step 9: Now, solve for x:

1. For cos(x) = (-1 + √2) / 2:

x = arccos((-1 + √2) / 2)

2. For cos(x) = (-1 - √2) / 2:

x = arccos((-1 - √2) / 2)

These are the two possible solutions for x. Keep in mind that trigonometric functions have periodicity, so there might be multiple solutions depending on the interval or range of x you are considering.