Question

Cos²x-7sin²x=-3sin2x​

Answer 1

Ответ:

$x1=arctg(\frac{6+\sqrt{8} }{14} )+\pi n\\x2=arctg(\frac{6-\sqrt{8} }{14})+\pi n$

Пошаговое объяснение:

$cos^{2}x-7sin^2x=-3sin2x\\ cos^2x-7sin^2x=-6sinxcosx\\cos^2x-7sin^2x+6sinxcosx=0 |:cos^2x\\1-7tg^2x+6tgx=0\\ tgx=t\\1-7t^2+6t=0\\7t^2-6t+1=0\\D=36-28=8\\t1=\frac{6+\sqrt{8} }{14}\\ t2=\frac{6-\sqrt{8} }{14} \\tgx= \frac{6+\sqrt{8} }{14} \\ x=arctg(\frac{6+\sqrt{8} }{14})+\pi n\\tgx = \frac{6-\sqrt{8} }{14} \\x= arctg(\frac{6-\sqrt{8} }{14}) + \pi n$

Советую вам пересмотреть условие, т.к. вы сами видите, какой получился ответ из-за не извлекаемого корня из дискриминанта.

Answer 2

To solve the trigonometric equation cos²x - 7sin²x = -3sin2x, we can use various trigonometric identities and algebraic manipulations to simplify and solve for x. Let's go step by step:

1. Start with the equation: cos²x - 7sin²x = -3sin2x

2. Use the trigonometric identity sin²x + cos²x = 1: cos²x = 1 - sin²x

3. Substitute the above identity into the equation: 1 - sin²x - 7sin²x = -3sin2x

4. Combine like terms on the left side: 1 - 8sin²x = -3sin2x

5. Move all terms to one side of the equation: 1 + 3sin2x - 8sin²x = 0

6. Factor out sin²x from the second and third terms: 1 + sin²x(3 - 8) = 0

7. Simplify the expression inside the parentheses: 1 - 5sin²x = 0

8. Divide both sides by -5: sin²x = -1/5

9. Since the range of sine squared is [0, 1], there are no real solutions for sin²x = -1/5. This means that the original equation has no real solutions for x.

In conclusion, the trigonometric equation cos²x - 7sin²x = -3sin2x has no real solutions.