Math / Trigonometry

Question11. Provide a trigonometric equation. Considering only the space between $x=0$ and $2 \pi$ , the equation must only have solutions at $x=1$ and $x=2$ . Explain your thought process and the work you did to create the equation. You may round decimal values to 3 places. [ 6 marks]

Studdy Solution

STEP 1

1. The trigonometric equation should have solutions only at $x = 1$ and $x = 2$ within the interval $0$ to $2\pi$ .
2. The equation can involve any standard trigonometric function(s).
3. Decimal values can be rounded to three decimal places.

STEP 2

1. Identify the trigonometric function to use.
2. Determine the form of the equation.
3. Adjust the equation to have solutions at specific points.
4. Verify the solutions.

STEP 3

Identify the trigonometric function to use. Since we need specific solutions at $x = 1$ and $x = 2$ , we can use the sine or cosine function, which have periodic characteristics and can be manipulated to have solutions at specific points.

STEP 4

Determine the form of the equation. We can start with a basic sine function and consider transformations:
$f(x) = \sin(kx + c)$
where $k$ and $c$ are constants to be determined.

STEP 5

Adjust the equation to have solutions at specific points. We want $f(x) = 0$ at $x = 1$ and $x = 2$ . Let's choose:
$f(x) = \sin(\pi(x - 1.5))$
This function will be zero at $x = 1$ and $x = 2$ because:
$\sin(\pi(1 - 1.5)) = \sin(-0.5\pi) = 0$ $\sin(\pi(2 - 1.5)) = \sin(0.5\pi) = 0$

STEP 6

Verify the solutions. Check that the function is zero only at $x = 1$ and $x = 2$ within the interval $0$ to $2\pi$ .
The function $\sin(\pi(x - 1.5))$ has a period of $2$ , so it will only be zero at $x = 1$ and $x = 2$ within the given interval.
The trigonometric equation is:
$\sin(\pi(x - 1.5)) = 0$

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Question11. Provide a trigonometric equation. Considering only the space between $x=0$ and $2 \pi$ , the equation must only have solutions at $x=1$ and $x=2$ . Explain your thought process and the work you did to create the equation. You may round decimal values to 3 places. [ 6 marks]

Studdy Solution

STEP 1

1. The trigonometric equation should have solutions only at $x = 1$ and $x = 2$ within the interval $0$ to $2\pi$ .
2. The equation can involve any standard trigonometric function(s).
3. Decimal values can be rounded to three decimal places.

STEP 2

1. Identify the trigonometric function to use.
2. Determine the form of the equation.
3. Adjust the equation to have solutions at specific points.
4. Verify the solutions.

STEP 3

Identify the trigonometric function to use. Since we need specific solutions at $x = 1$ and $x = 2$ , we can use the sine or cosine function, which have periodic characteristics and can be manipulated to have solutions at specific points.

STEP 4

Determine the form of the equation. We can start with a basic sine function and consider transformations:
$f(x) = \sin(kx + c)$
where $k$ and $c$ are constants to be determined.

STEP 5

Adjust the equation to have solutions at specific points. We want $f(x) = 0$ at $x = 1$ and $x = 2$ . Let's choose:
$f(x) = \sin(\pi(x - 1.5))$
This function will be zero at $x = 1$ and $x = 2$ because:
$\sin(\pi(1 - 1.5)) = \sin(-0.5\pi) = 0$ $\sin(\pi(2 - 1.5)) = \sin(0.5\pi) = 0$

STEP 6

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Studdy Solution

STEP 1

1. The trigonometric equation should have solutions only at x=1 x = 1 x=1 and x=2 x = 2 x=2 within the interval 0 0 0 to 2π 2\pi 2π.2. The equation can involve any standard trigonometric function(s).3. Decimal values can be rounded to three decimal places.

STEP 2

1. Identify the trigonometric function to use.2. Determine the form of the equation.3. Adjust the equation to have solutions at specific points.4. Verify the solutions.

STEP 3

Identify the trigonometric function to use. Since we need specific solutions at x=1 x = 1 x=1 and x=2 x = 2 x=2, we can use the sine or cosine function, which have periodic characteristics and can be manipulated to have solutions at specific points.

STEP 4

Determine the form of the equation. We can start with a basic sine function and consider transformations:f(x)=sin⁡(kx+c) f(x) = \sin(kx + c) f(x)=sin(kx+c)where k k k and c c c are constants to be determined.

STEP 5

STEP 6

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The trigonometric equation should have solutions only at $x = 1$ and $x = 2$ within the interval $0$ to $2\pi$ .
2. The equation can involve any standard trigonometric function(s).
3. Decimal values can be rounded to three decimal places.

1. Identify the trigonometric function to use.
2. Determine the form of the equation.
3. Adjust the equation to have solutions at specific points.
4. Verify the solutions.

Identify the trigonometric function to use. Since we need specific solutions at $x = 1$ and $x = 2$ , we can use the sine or cosine function, which have periodic characteristics and can be manipulated to have solutions at specific points.

Determine the form of the equation. We can start with a basic sine function and consider transformations:
$f(x) = \sin(kx + c)$
where $k$ and $c$ are constants to be determined.