Solve

Question

cot^2x+cosx=sin^2x?

Answer 1

The Function is Cot²x+cosx = sin²x

(cos²x/sin²x) + cosx - sin²x = 0

Multiply each side by sin²x

cos²x + sin²x * cosx - sin^4(x) = 0

cos²x + (1 - cos²x)cosx - (1 - cos²x)² = 0

cos²x + cosx - cos³x - (1 + Cos⁴x - 2cos²x) = 0

cos²x + cosx - cos³x - 1 - Cos⁴x + 2cos²x  = 0

- Cos⁴x - cos³x + 3cos²x + cosx - 1 = 0

Cos⁴x + cos³x - 3cos²x - cosx + 1 = 0

Let us ssume cosx = u therefore

u^4 + u³ - 3u² - u + 1 = 0

 

Solve the polynomial, we get the roots of the Polynomial as u = 0.47726, -0.73764, 1.35567 and -2.09529

Therefore cosx = 0.47726, -0.73764, 1.35567 and -2.09529

Since cosx cannot be greater than +1 and less than -1, so cosx = 0.47726, -0.73764

 

If cosx = 0.4776

x = 2nπ ± 61.49

For n = 0

x = 61.49 and -61.49

For n = 1

x = 298.51 and 421.49

 

If cosx = - 0.73764

x = 2nπ ± 137.53

For n = 0

x = 137.53 and -137.53

For n = 1

x = 497.53 and 222.47

 

Solutions of x are -137.49, -61.49, 61.49, 137.53, 222.47, 298.51 and so on.