Question

1)cos2x-cosx-1=0 2)sinx-1/cosx=0

Answer 1

Ответ:

1. cosx = cos2x-1

2. sinx = cosx+1

Объяснение:

1. cos2x-cosx-1=0

cos2x = cosx+1

cosx = cos2x-1

2. sinx-1/cosx=0

cosx = sinx-1

sinx = cosx+1

Answer 2

Let's solve each equation step by step:

1. cos(2x) - cos(x) - 1 = 0:

First, let's simplify the equation. We can use the identity cos(2x) = 2cos^2(x) - 1:

2cos^2(x) - 1 - cos(x) - 1 = 0

Rearranging the terms:

2cos^2(x) - cos(x) - 2 = 0

Now, we can factor the quadratic equation:

(2cos(x) + 1)(cos(x) - 2) = 0

Setting each factor equal to zero:

2cos(x) + 1 = 0 -> cos(x) = -1/2

cos(x) - 2 = 0 -> cos(x) = 2

To find the solutions, we need to consider the domain of the cosine function, which is from 0 to 2π (or 0 to 360 degrees).

For cos(x) = -1/2: In the given domain, x can take on two values: x = 120 degrees or x = 240 degrees

For cos(x) = 2: There are no solutions because the cosine function only ranges from -1 to 1.

Therefore, the solutions for the equation cos(2x) - cos(x) - 1 = 0 are: x = 120 degrees and x = 240 degrees.

2. sin(x) - (1/cos(x)) = 0:

Let's simplify the equation by multiplying through by cos(x):

cos(x) * sin(x) - 1 = 0

Using the identity sin(x) = cos(π/2 - x):

cos(x) * cos(π/2 - x) - 1 = 0

Expanding the equation:

cos^2(x) - cos(x)cos(π/2) - 1 = 0

cos(π/2) = 0, so the equation becomes:

cos^2(x) - 1 = 0

Factoring the quadratic equation:

(cos(x) - 1)(cos(x) + 1) = 0

Setting each factor equal to zero:

cos(x) - 1 = 0 -> cos(x) = 1

cos(x) + 1 = 0 -> cos(x) = -1

For cos(x) = 1: In the given domain, x can take on one value: x = 0 degrees

For cos(x) = -1: In the given domain, x can take on one value: x = 180 degrees

Therefore, the solutions for the equation sin(x) - (1/cos(x)) = 0 are: x = 0 degrees and x = 180 degrees.