Question

А)косинус х -2 синус х=0б)косинус в квадрате х +4 косинус х=0в)синус 3х - синус 7х=0г)7синус2х- 2 синус х=0 д)синус 2 х+10косинус в квадрате х=0е)5 косинус 2х-14косинус в квадрате х+8=0решите плииз .кто решит на счет деньги кину((((((((((((((((((((((

Answer 1

$1)\;\cos x -2\sin x=0\\ \cos x=2\sin x\\ctgx=2\\x=arcctg2\\ 2)\;\cos^2x +4\cos x=0\\\cos x(\cos x+4)=0\\\cos x=0\Rightrarrow x=\frac\pi2\\\cos x+4=0\Rightarrow\cos x=-4\;-\;pew.\;HET\\ 3)\;\sin3x-\sin7x=0\\\sin7x-\sin3x=0\\2\sin2x\cos5x=0\\\sin2x=0\Rightarrow2x=\pi n\Rightarrow x=\frac\pi2n$
$4)\;7\sin2x-2\sin x=0\\14\sin x\cosx-2\sinx=0\\2\sin x(7\cos x-1)=0\\\sin x=0\Rightarrow x=\pi n\\7\cos x-1=0\Rightarrow \cos x=\frac17\Rightarrow x=\arccos\frac17\\ 5)\;\sin2x+10\cos^2x=0\\2\sin x\cos x+10\cos^2x=0\\2\cos x(\sin x+5\cos x)=0\\\cos x=0\Rightarrow x=\frac\pi2+2\pi n\\\sin x+5\cos x=0\Rightarrow tgx=-5\Rightarrow x=arctg(-5)+\pi n$
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Answer 2

a) косинус х - 2 синус х = 0

To solve the equation cos(x) - 2sin(x) = 0, we can use the trigonometric identity sin^2(x) + cos^2(x) = 1. Rearranging the equation, we get:

cos(x) = 2sin(x)

Substituting cos(x) = √(1 - sin^2(x)), we have:

√(1 - sin^2(x)) = 2sin(x)

Squaring both sides of the equation, we get:

1 - sin^2(x) = 4sin^2(x)

Rearranging the equation, we have:

5sin^2(x) = 1

Dividing both sides by 5, we get:

sin^2(x) = 1/5

Taking the square root of both sides, we have:

sin(x) = ±√(1/5)

Taking the inverse sine (sin^(-1)) of both sides, we get:

x = sin^(-1)(±√(1/5))

The solutions for x will depend on the range of x that you are considering. The inverse sine function has multiple solutions within a given range. Please specify the range of x for a more precise answer.

b) косинус в квадрате х + 4 косинус х = 0

To solve the equation cos^2(x) + 4cos(x) = 0, we can factor out a common term:

cos(x)(cos(x) + 4) = 0

This equation will be satisfied if either cos(x) = 0 or cos(x) + 4 = 0.

For cos(x) = 0, the solutions are x = π/2 + nπ and x = 3π/2 + nπ, where n is an integer.

For cos(x) + 4 = 0, we have cos(x) = -4, which has no real solutions since the cosine function has a maximum value of 1 and a minimum value of -1.

Therefore, the solutions for the equation are x = π/2 + nπ and x = 3π/2 + nπ, where n is an integer.

c) синус 3х - синус 7х = 0

To solve the equation sin(3x) - sin(7x) = 0, we can use the trigonometric identity sin(a) - sin(b) = 2cos((a+b)/2)sin((a-b)/2). Applying this identity, we have:

2cos(5x/2)sin(-2x/2) = 0

Simplifying further, we get:

-2cos(5x/2)sin(x) = 0

This equation will be satisfied if either cos(5x/2) = 0 or sin(x) = 0.

For cos(5x/2) = 0, the solutions are x = π/5 + nπ, where n is an integer.

For sin(x) = 0, the solutions are x = nπ, where n is an integer.

Therefore, the solutions for the equation are x = π/5 + nπ and x = nπ, where n is an integer.

d) 7синус2х - 2 синус х = 0

To solve the equation 7sin(2x) - 2sin(x) = 0, we can factor out a common term:

sin(x)(7sin(x) - 2) = 0

This equation will be satisfied if either sin(x) = 0 or 7sin(x) - 2 = 0.

For sin(x) = 0, the solutions are x = nπ, where n is an integer.

For 7sin(x) - 2 = 0, we have sin(x) = 2/7, which has no real solutions since the sine function has a maximum value of 1 and a minimum value of -1.

Therefore, the solutions for the equation are x = nπ, where n is an integer.

e) синус 2 х + 10косинус в квадрате х = 0

To solve the equation sin(2x) + 10cos^2(x) = 0, we can use the trigonometric identity sin(2a) = 2sin(a)cos(a). Applying this identity, we have:

2sin(x)cos(x) + 10cos^2(x) = 0

Factoring out a common term, we get:

cos(x)(2sin(x) + 10cos(x)) = 0

This equation will be satisfied if either cos(x) = 0 or 2sin(x) + 10cos(x) = 0.

For cos(x) = 0, the solutions are x = π/2 + nπ and x = 3π/2 + nπ, where n is an integer.

For 2sin(x) + 10cos(x) = 0, we can divide both sides by 2 to simplify the equation:

sin(x) + 5cos(x) = 0

This equation does not have a simple algebraic solution. However, it can be solved numerically or graphically.

Therefore, the solutions for the equation are x = π/2 + nπ and x = 3π/2 + nπ, where n is an integer, and the equation sin(x) + 5cos(x) = 0 does not have a simple algebraic solution.

f) 5 + косинус 2х - 14косинус в квадрате х + 8 = 0

To solve the equation 5 + cos(2x) - 14cos^2(x) + 8 = 0, we can simplify it by combining like terms:

cos(2x) - 14cos^2(x) + 13 = 0

This equation does not have a simple algebraic solution. However, it can be solved numerically or graphically.

Therefore, the equation cos(2x) - 14cos^2(x) + 13 = 0 does not have a simple algebraic solution.