Question

2cos^2x+sin2x-8cos2x=0

Answer 1

2Сos^2x + 2SinxCosx - 8(Cos^2x - Sin^2x) = 0
2Cos^2x + 2SinxCosx - 8Cos^2x + 8Sin^2x = 0
8Sin^2x + 2SinxCosx -6Cos^2x = 0 | : Cos^2x
8tg^2x +2tgx -6 = 0
4tg^2x + tgx -3 = 0
Решаем как квадратное.
D=49
a)tgx = (-1 +7)/8 = 6/8 =3/4
x = arctg(3/4) +πk. k Є Z
б) tg x = (-1-8)/8 = -1
x = - π/4 + πk. k Є Z

Answer 2

To solve the equation 2cos^2(x) + sin(2x) - 8cos(2x) = 0, you can use trigonometric identities and algebraic manipulation to simplify and solve for x. Let's break it down step by step:

1. Start by using the double angle identity for cosine: cos(2x) = 2cos^2(x) - 1. Replace cos(2x) in the equation:

2cos^2(x) + sin(2x) - 8(2cos^2(x) - 1) = 0

2. Expand and simplify the equation:

2cos^2(x) + sin(2x) - 16cos^2(x) + 8 = 0

3. Combine like terms:

-14cos^2(x) + sin(2x) + 8 = 0

4. Now, use the double angle identity for sine: sin(2x) = 2sin(x)cos(x). Replace sin(2x) in the equation:

-14cos^2(x) + 2sin(x)cos(x) + 8 = 0

5. Factor out common terms:

2(cos(x))(7cos(x) - 4sin(x)) + 8 = 0

6. Divide the entire equation by 2 to simplify further:

(cos(x))(7cos(x) - 4sin(x)) + 4 = 0

7. Now, apply the Pythagorean identity for sine and cosine: sin(x) = √(1 - cos^2(x)). Replace sin(x) in the equation:

(cos(x))(7cos(x) - 4√(1 - cos^2(x))) + 4 = 0

8. Distribute and simplify:

7cos^2(x) - 4√(1 - cos^2(x))cos(x) + 4 = 0

9. Now, let's make a substitution to simplify this equation. Let y = cos(x). The equation becomes:

7y^2 - 4√(1 - y^2)y + 4 = 0

10. This is a quadratic equation in y. You can solve it using the quadratic formula:

y = [4√(1 - y^2) ± √(16(1 - y^2) - 4(7)(4))] / (2(7))

11. Simplify the equation further:

y = [4√(1 - y^2) ± √(16 - 28(1 - y^2))] / 14

12. Now, solve for y and then substitute back to find the solutions for x.

It's worth noting that this equation may have multiple solutions depending on the value of y, and you may need to use trigonometric identities and solve for x in specific intervals to find the solutions in the desired range.