Question

Cos2x-2sinx×cos2x=0

Answer 1

Cos 2x(1-2sinx)=0 cos 2x=0 2x=пи\2+пи*n x= пи\4+(пи\2)n 2sinx=1 sinx=1\2 x= минус 1 в степени n умноженное на пи\6 + пи умноженное на n

Answer 2

I can help you solve this trigonometric equation. First, let's rewrite cos2x using the double angle formula:

$$cos2x = cos^2x - sin^2x$$

Then, we can substitute this into the original equation and simplify:

$$cos^2x - sin^2x - 2sinx \times cos^2x - sin^2x = 0$$

$$cos^2x - 3sin^2x - 2sinx \times cos^2x = 0$$

Next, we can factor out cos^2x from the equation and get:

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

This equation is true when either cos^2x = 0 or 1 - 3sinx - 2sinx = 0. Let's solve for each case separately.

If cos^2x = 0, then cosx = 0. This happens when x is an odd multiple of $\frac{\pi}{2}$, or:

$$x = \frac{(2n + 1)\pi}{2}, n \in \mathbb{Z}$$

If 1 - 3sinx - 2sinx = 0, then we can use the quadratic formula to solve for sinx:

$$sinx = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times (-2)}}{2 \times 1}$$

$$sinx = \frac{3 \pm \sqrt{17}}{4}$$

However, since the sine function has a range of [-1, 1], only the negative root is valid:

$$sinx = \frac{3 - \sqrt{17}}{4}$$

To find the value of x, we can use the inverse sine function and the periodicity of sine:

$$x = sin^{-1}(\frac{3 - \sqrt{17}}{4}) + 2\pi n, n \in \mathbb{Z}$$

Therefore, the general solution of the equation is:

$$x = \frac{(2n + 1)\pi}{2}, sin^{-1}(\frac{3 - \sqrt{17}}{4}) + 2\pi n, n \in \mathbb{Z}$$

You can check your answer using this [online calculator](https://www.symbolab.com/solver/trigonometric-equation-calculator/cos2x%2Bsinx%3D0). I hope this helps you understand how to solve trigonometric equations. If you have any other questions, feel free to ask me.