Trig Equations
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Question 1 of 3
1. Question
Given that `0≥theta≥360`, solve:`cos theta=- 1/sqrt2`- `theta=` (135, 225)`°,` (225, 135)`°`
Hint
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Well Done!
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Trigonometric Functions
$$\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\tan\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$
Positive Values in the Unit Circle
Quadrant I: AllQuadrant II: Sine onlyQuadrant III: Tangent onlyQuadrant IV: Cosine onlyExact Triangle Ratios
The acute angle `(theta’)` is an angle less than `90°` and is relative to the horizontal axis.First, identify which quadrant `theta` may lie.Since the given cosine value is negative, `theta` may be on Quadrant II or III.
Next, get the acute angle (`theta’`) for `theta`.We can do this by matching the value of `cos theta` to an Exact Triangle.$$\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}=-\frac{\color{#00880A}{1}}{\color{#007DDC}{\sqrt2}}$$
Since `1` is the adjacent side and `sqrt2` is the hypotenuse, `theta’=45°`Now, find the corresponding angle of the acute angle in both Quadrant II and III.Quadrant II:`180°-45°=``135°`
Quadrant III:`180°+45°=``225°`
Therefore, `theta=135°,225°``theta=135°,225°` -
Question 2 of 3
2. Question
Given that `0≥theta≥360`, solve:`sin theta=(sqrt3)/2`- `theta=` (60, 120)`°,` (120, 60)`°`
Hint
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Fantastic!
Incorrect
Trigonometric Functions
$$\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\tan\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$
Positive Values in the Unit Circle
Quadrant I: AllQuadrant II: Sine onlyQuadrant III: Tangent onlyQuadrant IV: Cosine onlyExact Triangle Ratios
The acute angle `(theta’)` is an angle less than `90°` and is relative to the horizontal axis.First, identify which quadrant `theta` may lie.Since the given sine value is positive, `theta` may be on Quadrant I or II.
Next, get the acute angle (`theta’`) for `theta`.We can do this by matching the value of `sin theta` to an Exact Triangle.$$\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}=\frac{\color{#9a00c7}{\sqrt3}}{\color{#007DDC}{2}}$$
Since `sqrt3` is the opposite side and `2` is the hypotenuse, `theta’=60°`Now, find the corresponding angle of the acute angle in both Quadrant I and II.Quadrant I:`60°`
Quadrant II:`180°-60°=``120°`
Therefore, `theta=60°,120°``theta=60°,120°` -
Question 3 of 3
3. Question
Given that `0≥x≥360`, solve:`tan x=-1`- `x=` (135, 315)`°,` (135, 315)`°`
Hint
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Great Work!
Incorrect
Trigonometric Functions
$$\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}$$$$\tan\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}$$Positive Values in the Unit Circle
Quadrant I: AllQuadrant II: Sine onlyQuadrant III: Tangent onlyQuadrant IV: Cosine onlyExact Triangle Ratios
The acute angle `(theta’)` is an angle less than `90°` and is relative to the horizontal axis.First, identify which quadrant `x` may lie.Since the given tangent value is negative, `x` may be on Quadrant II or IV.
Next, get the acute angle for `x`.We can do this by matching the value of `tan x` to an Exact Triangle.$$\tan x=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}=-\frac{\color{#9a00c7}{1}}{\color{#00880A}{1}}$$
Since `1` is the opposite side and `1` is the adjacent, the acute angle is `45°`Now, find the corresponding angle of the acute angle in both Quadrant II and IV.Quadrant II:`180°-45°=``135°`
Quadrant IV:`360°-45°=``315°`
Therefore, `x=135°,315°``x=135°,315°`
Quizzes
- Converting Angle Measures 1
- Converting Angle Measures 2
- Converting Angle Measures 3
- Finding the Central Angle in a Circle
- Finding Areas in a Circle
- Values on the Unit Circle
- Finding Missing Angles Using the Unit Circle
- Trigonometric Ratios in the Unit Circle
- Trig Exact Values 1
- Trig Exact Values 2
- Trig Equations
- Derivative of a Trigonometric Function 1
- Derivative of a Trigonometric Function 2
- Derivative of a Trigonometric Function 3
- Applications of Differentiation
- Integral of a Trigonometric Function 1
- Integral of a Trigonometric Function 2
- Applications of Integration
- Graphing Trigonometric Functions 1
- Graphing Trigonometric Functions 2
- Graphing Trigonometric Functions 3
- Graphing Trigonometric Functions 4