Question

1).cos x + sin(Pi/2 -x) + cos (Pi + x)=0 2). cos(Pi+x)=cos Pi/6 3). 7cos(x- 3pi/2)+5sin x+1=0 4). cos2x+cosx=-sin2x

Answer 1

1
cosx+cosx-cosx=0
cosx=0
x=π/2+πn,n∈z
2
cos(Pi+x)=cos Pi/6
-cosx=√3/2
cosx=-√3/2
x=+-5π/6+2πn,n∈z
3
7cos(x- 3pi/2)+5sin x+1=0
-7sinx+5sinx=-1
-2sinx=-1
sinx=1/2
x=(-1)^n*π/6+πn,n∈z
4
cos2x+cosx=-sin2x

Answer 2

1) Equation: cos(x) + sin(Pi/2 - x) + cos(Pi + x) = 0

Let's simplify the equation step by step:

First, we know that sin(Pi/2 - x) = cos(x). So, substituting this in the equation, we get:

cos(x) + cos(x) + cos(Pi + x) = 0

Now, we can simplify further by using the trigonometric identity cos(Pi + x) = -cos(x). Substituting this in the equation, we get:

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

Simplifying, we have:

cos(x) = 0

To find the values of x that satisfy this equation, we need to determine the solutions for which cos(x) = 0. In the trigonometric unit circle, cos(x) = 0 at specific angles. These angles occur at x = Pi/2 + n * Pi, where n is an integer.

So, the solutions for the equation cos(x) + sin(Pi/2 - x) + cos(Pi + x) = 0 are:

x = Pi/2 + n * Pi, where n is an integer.

2) Equation: cos(Pi + x) = cos(Pi/6)

Again, we can use the trigonometric identity cos(Pi + x) = -cos(x). Substituting this in the equation, we get:

-cos(x) = cos(Pi/6)

To find the values of x that satisfy this equation, we need to determine the solutions for which -cos(x) = cos(Pi/6).

Using the unit circle, we can find the solutions. At x = Pi/6, cos(Pi/6) = sqrt(3)/2. So, the equation becomes:

-cos(x) = sqrt(3)/2

Multiplying both sides by -1, we get:

cos(x) = -sqrt(3)/2

To find the angles where cos(x) = -sqrt(3)/2, we can refer to the unit circle. These angles occur at x = 2Pi/3 + 2n * Pi or x = 4Pi/3 + 2n * Pi, where n is an integer.

So, the solutions for the equation cos(Pi + x) = cos(Pi/6) are:

x = 2Pi/3 + 2n * Pi, or x = 4Pi/3 + 2n * Pi, where n is an integer.

3) Equation: 7cos(x - 3Pi/2) + 5sin(x) + 1 = 0

Let's simplify the equation step by step:

First, we know that cos(x - 3Pi/2) = sin(x). So, substituting this in the equation, we get:

7sin(x) + 5sin(x) + 1 = 0

Simplifying further, we have:

12sin(x) = -1

Dividing both sides by 12, we get:

sin(x) = -1/12

To find the angles where sin(x) = -1/12, we can refer to the unit circle. These angles occur at x = -sin^(-1)(1/12) + 2n * Pi or x = Pi + sin^(-1)(1/12) + 2n * Pi, where n is an integer.

So, the solutions for the equation 7cos(x - 3Pi/2) + 5sin(x) + 1 = 0 are:

x = -sin^(-1)(1/12) + 2n * Pi, or x = Pi + sin^(-1)(1/12) + 2n * Pi, where n is an integer.

4) Equation: cos(2x) + cos(x) = -sin(2x)

To simplify this equation, we can use the trigonometric identity cos(2x) = 2cos^2(x) - 1 and sin(2x) = 2sin(x)cos(x). Substituting these in the equation, we get:

2cos^2(x) - 1 + cos(x) = -2sin(x)cos(x)

Rearranging the terms, we have:

2cos^2(x) + cos(x) + 2sin(x)cos(x) + 1 = 0

Now, let's simplify further:

2cos^2(x) + cos(x) + 2sin(x)cos(x) + 1 = 0

Multiplying each term by 2, we get:

4cos^2(x) + 2cos(x) + 4sin(x)cos(x) + 2 = 0

Rearranging the terms, we have:

4cos^2(x) + 4sin(x)cos(x) + 2cos(x) + 2 = 0

Factoring out a common term, we get:

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

Dividing both sides by 2, we have:

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

To find the values of x that satisfy this equation, we need to determine the solutions for which cos(x)(2cos(x) + 2sin(x) + 1) + 1 = 0. Unfortunately, this equation does not have simple analytical solutions. You would need to use numerical methods or approximation techniques to find the values of x that satisfy this equation.