Question

5cos²x+7cosx-6=08cos²x-10sinx-11=0

Answer 1

$5cos^2x+7cosx-6=0\\\\cosx=t,\; \; 5t^2+7t-6=0\\\\D=49+4*5*6=169\\\\t_1=\frac{-7-13}{10}=-2,\; \; t_2=\frac{6}{10}=\frac{3}{5}\\\\cosx=-2\ \textless \ -1\; \to \; net\; resheniya\\\\cosx=\frac{3}{5},\; x=\pm arccos\frac{3}{5}+2\pi n,\; n\in Z$


$8cos^2x-10sinx-11=0\\\\8(1-sin^2x)-10sinx-11=0\\\\8sin^2x+10sinx+2=0\\\\t=sinx,\; 8t^2+10t+2=0;4t^2+5t+1=0\\\\D=25-16=9\\\\t_1=\frac{-5-3}{8}=-1,\; \; t_2=\frac{-5+3}{8}=-\frac{1}{4}\\\\sinx=-1,\; \; x=-\frac{\pi}{2}+2\pi n,\; n\in Z\\\\sinx=-\frac{1}{4},\\\\ x=(-1)^{k}\cdot arcsin(-\frac{1}{4})+\pi k=(-1)^{k+1}\cdot arccos\frac{1}{4}+\pi k,\; k\in Z$

Answer 2

Решение во вложении...........

Answer 3

To solve the system of equations:

1. 5cos²x + 7cosx - 6 = 0
2. 8cos²x - 10sinx - 11 = 0

We can start by using the trigonometric identity for sin²x:

sin²x = 1 - cos²x

Now, let's substitute this identity into the second equation:

8cos²x - 10(1 - cos²x) - 11 = 0

Now, simplify and solve for cos²x:

8cos²x - 10 + 10cos²x - 11 = 0

Combine like terms:

18cos²x - 21 = 0

Now, isolate cos²x:

18cos²x = 21

cos²x = 21/18

cos²x = 7/6

Now, take the square root of both sides, but remember that cos²x is always positive, so we only consider the positive square root:

cosx = √(7/6)

cosx ≈ ±0.9354

Now, let's solve the first equation, 5cos²x + 7cosx - 6 = 0, using the value of cosx we found:

5(7/6) + 7cosx - 6 = 0

(35/6) + 7cosx - 6 = 0

7cosx = 6 - (35/6)

7cosx = (36/6) - (35/6)

7cosx = (1/6)

cosx = (1/6) / 7

cosx ≈ 1/42

So, the solutions for x are approximately:

1. x ≈ arccos(√(7/6)) or x ≈ -arccos(√(7/6))
2. x ≈ arccos(1/42) or x ≈ -arccos(1/42)

These are the approximate values for x that satisfy the given system of equations.