Question

4cos²x-2sin²x-5cosx-4=0

Answer 1

$4Cos^{2}x-2Sin^{2}x-5Cosx-4=0\\\\4Cos^{2}x-2(1-Cos^{2}x)-5Cosx-4=0\\\\4Cos^{2}x-2+2Cos^{2}x-5Cosx-4=0\\\\6Cos^{2}x-5Cosx-6=0\\\\Cosx=m,-1 \leq m \leq 1 \\\\6m^{2}-5m-6=0\\\\D=(-5)^{2}-4*6*(-6)=25+144=169=13^{2}\\\\m_{1}=\frac{5-13}{12}=-\frac{8}{12}=-\frac{2}{3} \\\\m_{2}=\frac{5+13}{12}=\frac{18}{12}=1,5>1-neyd\\\\Cosx=-\frac{2}{3}\\\\x=\pm arc Cos(-\frac{2}{3})+2\pi n,n\in Z$

Answer 2

To solve the equation 4cos²x - 2sin²x - 5cosx - 4 = 0, we can use some trigonometric identities to simplify it and then solve for the unknown variable x.

First, we'll use the identity: sin²x + cos²x = 1

Now, rewrite the equation with sin²x in terms of cos²x:

4cos²x - 2(1 - cos²x) - 5cosx - 4 = 0

Simplify further:

4cos²x - 2 + 2cos²x - 5cosx - 4 = 0

Combine like terms:

6cos²x - 5cosx - 6 = 0

Now, let's solve this quadratic equation for cosx. We can either factor it or use the quadratic formula. Let's use the quadratic formula:

The quadratic formula is given by: x = (-b ± √(b² - 4ac)) / 2a

In this equation, a = 6, b = -5, and c = -6.

Now, plug in the values:

cosx = [5 ± √((-5)² - 4 * 6 * (-6))] / (2 * 6)

cosx = [5 ± √(25 + 144)] / 12

cosx = [5 ± √169] / 12

cosx = (5 ± 13) / 12

Now, we have two possible solutions for cosx:

1. cosx = (5 + 13) / 12 = 18 / 12 = 3 / 2 (Not a valid solution, as the cosine function is bound between -1 and 1)

2. cosx = (5 - 13) / 12 = -8 / 12 = -2 / 3

Now, we can find the corresponding values of sinx using the identity sin²x + cos²x = 1:

sin²x = 1 - cos²x sin²x = 1 - (-2/3)² sin²x = 1 - 4/9 sin²x = 5/9

sinx = ±√(5/9) = ±√5/3

So, the two possible solutions for the equation are:

1. x = arcsin(√5/3) + 2πn
2. x = π - arcsin(√5/3) + 2πn

where n is an integer representing the number of full circles or revolutions made around the unit circle.

Note: Some of these solutions may fall outside the typical range of angles, i.e., [-π, π]. So, we use the general solutions with n to account for all possible solutions.