How many values of θ will satisfy the equation (sin2θ - 4 sin θ + 3) (4 - cos2θ + 4 sin θ) = 0, where 0 < θ < \(\frac{\pi}{2}\) ?
None
Solving Trigonometric Equations for Theta in an Interval
The problem asks us to find the number of values of \(\theta\) that satisfy the given equation in the specific interval \(0 \lt \theta \lt \frac{\pi}{2}\). The equation is:
\( (\sin^2\theta - 4 \sin\theta + 3) (4 - \cos^2\theta + 4 \sin\theta) = 0 \)
For a product of two factors to be zero, at least one of the factors must be zero. So, we need to analyze two separate cases:
- The first factor is zero: \(\sin^2\theta - 4 \sin\theta + 3 = 0\)
- The second factor is zero: \(4 - \cos^2\theta + 4 \sin\theta = 0\)
We will solve each case and then check if the resulting values of \(\theta\) fall within the given interval \(0 \lt \theta \lt \frac{\pi}{2}\). Remember, the interval \(0 \lt \theta \lt \frac{\pi}{2}\) means \(\theta\) must be strictly greater than 0 and strictly less than \(\frac{\pi}{2}\). This is the first quadrant, excluding the axes.
Analyzing the First Factor: \(\sin^2\theta - 4 \sin\theta + 3 = 0\)
This equation looks like a quadratic equation if we consider \(\sin\theta\) as the variable. Let \(x = \sin\theta\). The equation becomes:
\( x^2 - 4x + 3 = 0 \)
We can factor this quadratic equation:
\( (x - 1)(x - 3) = 0 \)
This gives us two possible values for \(x\):
- \(x - 1 = 0 \implies x = 1\)
- \(x - 3 = 0 \implies x = 3\)
Substituting back \(x = \sin\theta\), we get:
- \(\sin\theta = 1\)
- \(\sin\theta = 3\)
Now, let's evaluate these possibilities for \(\sin\theta\). We know that the sine function has a range of \([-1, 1]\). This means the value of \(\sin\theta\) must be between -1 and 1, inclusive.
- \(\sin\theta = 1\): This value is within the valid range \([-1, 1]\). The general solution for \(\sin\theta = 1\) is \(\theta = \frac{\pi}{2} + 2n\pi\), where \(n\) is an integer. In the interval \([0, 2\pi]\), the only solution is \(\theta = \frac{\pi}{2}\). However, the given interval is \((0, \frac{\pi}{2})\), which means \(\theta\) must be strictly less than \(\frac{\pi}{2}\). Therefore, \(\theta = \frac{\pi}{2}\) is not included in the interval \((0, \frac{\pi}{2})\). So, \(\sin\theta = 1\) yields no solutions in the given interval.
- \(\sin\theta = 3\): This value is outside the valid range \([-1, 1]\) because \(3 > 1\). Therefore, \(\sin\theta = 3\) has no real solutions for \(\theta\).
From the first factor, there are no values of \(\theta\) in the interval \((0, \frac{\pi}{2})\) that satisfy \(\sin^2\theta - 4 \sin\theta + 3 = 0\).
Analyzing the Second Factor: \(4 - \cos^2\theta + 4 \sin\theta = 0\)
This equation involves both \(\sin\theta\) and \(\cos^2\theta\). We can use the trigonometric identity \(\cos^2\theta = 1 - \sin^2\theta\) to express the equation entirely in terms of \(\sin\theta\). Substituting the identity:
\( 4 - (1 - \sin^2\theta) + 4 \sin\theta = 0 \)
Simplify the equation:
\( 4 - 1 + \sin^2\theta + 4 \sin\theta = 0 \)
\( \sin^2\theta + 4 \sin\theta + 3 = 0 \)
Again, this is a quadratic equation in \(\sin\theta\). Let \(y = \sin\theta\). The equation is:
\( y^2 + 4y + 3 = 0 \)
We can factor this quadratic equation:
\( (y + 1)(y + 3) = 0 \)
This gives us two possible values for \(y\):
- \(y + 1 = 0 \implies y = -1\)
- \(y + 3 = 0 \implies y = -3\)
Substituting back \(y = \sin\theta\), we get:
- \(\sin\theta = -1\)
- \(\sin\theta = -3\)
Let's evaluate these possibilities for \(\sin\theta\) considering the valid range \([-1, 1]\).
- \(\sin\theta = -1\): This value is within the valid range \([-1, 1]\). The general solution for \(\sin\theta = -1\) is \(\theta = \frac{3\pi}{2} + 2n\pi\), where \(n\) is an integer. In the interval \([0, 2\pi]\), the only solution is \(\theta = \frac{3\pi}{2}\). The given interval for \(\theta\) is \((0, \frac{\pi}{2})\). This interval is in the first quadrant, where the sine function is positive. The value \(\theta = \frac{3\pi}{2}\) is in the third quadrant and is not in the interval \((0, \frac{\pi}{2})\). So, \(\sin\theta = -1\) yields no solutions in the given interval.
- \(\sin\theta = -3\): This value is outside the valid range \([-1, 1]\) because \(-3 < -1\). Therefore, \(\sin\theta = -3\) has no real solutions for \(\theta\).
From the second factor, there are no values of \(\theta\) in the interval \((0, \frac{\pi}{2})\) that satisfy \(4 - \cos^2\theta + 4 \sin\theta = 0\).
Conclusion on Number of Solutions
We analyzed both factors of the original trigonometric equation. The first factor led to \(\sin\theta = 1\) or \(\sin\theta = 3\). Neither of these resulted in a valid \(\theta\) in the interval \((0, \frac{\pi}{2})\). The second factor led to \(\sin\theta = -1\) or \(\sin\theta = -3\). Neither of these resulted in a valid \(\theta\) in the interval \((0, \frac{\pi}{2})\).
Since neither part of the equation is satisfied by any \(\theta\) in the specified interval, there are no values of \(\theta\) that satisfy the original equation in the range \(0 \lt \theta \lt \frac{\pi}{2}\).
Therefore, the number of values of \(\theta\) that satisfy the equation in the given interval is none.
Trigonometric Equation Solution Review
Let's quickly review the key steps and results:
| Factor | Equation | Solutions for \(\sin\theta\) | Valid \(\sin\theta\) in \([-1, 1]\) | \(\theta\) in \((0, \frac{\pi}{2})\) |
|---|---|---|---|---|
| First Factor | \(\sin^2\theta - 4 \sin\theta + 3 = 0\) | \(\sin\theta = 1\), \(\sin\theta = 3\) | \(\sin\theta = 1\) (Valid), \(\sin\theta = 3\) (Invalid) | For \(\sin\theta = 1\), \(\theta = \frac{\pi}{2}\) (Not in interval). No solutions from \(\sin\theta=3\). |
| Second Factor | \(4 - \cos^2\theta + 4 \sin\theta = 0\) becomes \(\sin^2\theta + 4 \sin\theta + 3 = 0\) |
\(\sin\theta = -1\), \(\sin\theta = -3\) | \(\sin\theta = -1\) (Valid), \(\sin\theta = -3\) (Invalid) | For \(\sin\theta = -1\), \(\theta = \frac{3\pi}{2}\) (Not in interval). No solutions from \(\sin\theta=-3\). |
Revision Table: Key Trigonometry Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Trigonometric Equation | An equation involving trigonometric functions of variables. | We solved a complex trigonometric equation. |
| Solving by Factoring | If \(A \times B = 0\), then \(A=0\) or \(B=0\). | Used to break the original equation into two simpler equations. |
| Quadratic in \(\sin\theta\) | Equations that can be written in the form \(a(\sin\theta)^2 + b(\sin\theta) + c = 0\). | Both factored equations resulted in quadratics in terms of \(\sin\theta\). |
| Range of \(\sin\theta\) | The values \(\sin\theta\) can take are between -1 and 1, i.e., \([-1, 1]\). | Used to determine if the calculated values for \(\sin\theta\) are possible. |
| Trigonometric Identities | Equations that are true for all values of the variables for which the functions are defined, e.g., \(\sin^2\theta + \cos^2\theta = 1\). | Used \(\cos^2\theta = 1 - \sin^2\theta\) to simplify the second factor. |
| Solving for \(\theta\) | Finding the angle(s) that satisfy a trigonometric equation. | We solved for \(\sin\theta\) and then considered the corresponding \(\theta\) values. |
| Angle Interval | A specified range of values for the angle \(\theta\). | The solutions must fall within the given open interval \((0, \frac{\pi}{2})\). |
Additional Information: Understanding Angle Intervals
When solving trigonometric equations, the interval for the angle \(\theta\) is very important. Solutions repeat periodically, but the interval restricts the solutions we are looking for. The interval \(0 \lt \theta \lt \frac{\pi}{2}\) corresponds to the first quadrant of the unit circle, excluding the positive x-axis (\(\theta=0\)) and the positive y-axis (\(\theta=\frac{\pi}{2}\)).
- In the interval \((0, \frac{\pi}{2})\), the sine function \(\sin\theta\) is strictly positive (\(0 \lt \sin\theta \lt 1\)).
- The value \(\sin\theta = 1\) occurs at \(\theta = \frac{\pi}{2}\), which is not included in the open interval \((0, \frac{\pi}{2})\).
- The value \(\sin\theta = -1\) occurs at \(\theta = \frac{3\pi}{2}\), which is far outside the interval \((0, \frac{\pi}{2})\).
- Values like \(\sin\theta = 3\) or \(\sin\theta = -3\) are impossible for any real angle \(\theta\).
By carefully considering the valid range of \(\sin\theta\) and the specified interval for \(\theta\), we could determine that none of the potential solutions for \(\sin\theta\) from either factor of the original equation lead to a valid \(\theta\) within \(0 \lt \theta \lt \frac{\pi}{2}\).
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