Trigonometric Equation: Solve 2 Sin^2(theta) + 7 Cos(theta) - 5 = 0
Hey math whizzes! Today, we're diving deep into the world of trigonometry to tackle a rather gnarly-looking equation: $2 \sin^2(\theta) + 7 \cos(\theta) - 5 = 0$. Our mission, should we choose to accept it (and we totally do!), is to find all the values of $\theta$ that satisfy this equation within the interval $0 \ \textless \ \theta \ \textless \ 2\pi$. Remember, we need our final answers in radians, so ditch those degrees!
This problem is a classic example of how we can use trigonometric identities to simplify complex equations. The key here is to express everything in terms of a single trigonometric function. We've got both sine and cosine terms, which is a bit of a red flag. The Pythagorean identity, $\sin^2(\theta) + \cos^2(\theta) = 1$, is our best friend for situations like this. We can rearrange it to get $\sin^2(\theta) = 1 - \cos^2(\theta)$. This is super handy because it allows us to substitute $1 - \cos^2(\theta)$ wherever we see $\sin^2(\theta)$, effectively transforming our equation into one solely in terms of cosine.
So, let's do that substitution. Replace $2 \sin^2(\theta)$ with $2(1 - \cos^2(\theta))$. Our equation now looks like this: $2(1 - \cos^2(\theta)) + 7 \cos(\theta) - 5 = 0$. Time to get rid of those parentheses by distributing the 2: $2 - 2\cos^2(\theta) + 7 \cos(\theta) - 5 = 0$. Now, let's combine the constant terms ($2$ and $-5$) to get $-3$. The equation becomes: $-2\cos^2(\theta) + 7 \cos(\theta) - 3 = 0$.
Looking at this, it strongly resembles a quadratic equation. If we let $x = \cos(\theta)$, then our equation transforms into $-2x^2 + 7x - 3 = 0$. See? It's just a quadratic in disguise! To make it even cleaner, let's multiply the entire equation by $-1$ to make the leading coefficient positive: $2x^2 - 7x + 3 = 0$. This is a standard quadratic equation that we can solve using factoring or the quadratic formula. Factoring often feels more satisfying, so let's give that a whirl.
We're looking for two numbers that multiply to $ac = (2)(3) = 6$ and add up to $b = -7$. The numbers $-1$ and $-6$ fit the bill, since $-1 \times -6 = 6$ and $-1 + (-6) = -7$. Now, we rewrite the middle term $-7x$ using these two numbers: $2x^2 - x - 6x + 3 = 0$. Next, we factor by grouping. Group the first two terms and the last two terms: $(2x^2 - x) + (-6x + 3) = 0$. Factor out the greatest common factor from each group: $x(2x - 1) - 3(2x - 1) = 0$. Notice that we have a common binomial factor $(2x - 1)$. Factor that out: $(2x - 1)(x - 3) = 0$.
Now, for the magic to happen, either $(2x - 1) = 0$ or $(x - 3) = 0$. Solving the first one, $2x = 1$, which gives us $x = 1/2$. Solving the second one, $x = 3$.
Remember, we made the substitution $x = \cos(\theta)$ earlier. So, we now have two possibilities for $\cos(\theta)$:
- $\cos(\theta) = 1/2$
- $\cos(\theta) = 3$
Let's analyze these two possibilities. For the first one, $\cos(\theta) = 1/2$, we know that the cosine function has values between $-1$ and $1$. Since $1/2$ is within this range, there are solutions for $\theta$. We're looking for angles in the interval $0 \ \textless \ \theta \ \textless \ 2\pi$ where the cosine is positive. This occurs in the first and fourth quadrants. The reference angle (the acute angle made with the x-axis) for which the cosine is $1/2$ is $\frac\pi}{3}$ radians. So, in the first quadrant, $\theta = \frac{\pi}{3}$. In the fourth quadrant, the angle is $2\pi - \frac{\pi}{3} = \frac{6\pi - \pi}{3} = \frac{5\pi}{3}$. Both of these angles are within our specified range. So, we have found two potential solutions{3}$ and $\theta = \frac{5\pi}{3}$.
Now, for the second possibility, $\cos(\theta) = 3$. As we just mentioned, the cosine function can only output values between $-1$ and $1$ (inclusive). Since $3$ is outside this range, there are no real solutions for $\theta$ that satisfy $\cos(\theta) = 3$. So, we can discard this case entirely. It's like trying to fit a square peg in a round hole – it just doesn't work!
Therefore, the only solutions to our original trigonometric equation $2 \sin^2(\theta) + 7 \cos(\theta) - 5 = 0$ within the interval $0 \ \textless \ \theta \ \textless \ 2\pi$ are $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$. We've successfully navigated the twists and turns of trigonometric identities and quadratic equations to arrive at our final answers. Pretty neat, huh guys?
Key Takeaways:
- Pythagorean Identity is Your Bestie: Whenever you see both $\sin^2(\theta)$ and $\cos^2(\theta)$ (or just $\sin^2(\theta)$ and $\cos(\theta)$, or vice versa), the identity $\sin^2(\theta) + \cos^2(\theta) = 1$ is your go-to for simplifying.
- Transform to Quadratics: Many trigonometric equations can be rewritten as quadratic equations by using substitutions. Once you have a quadratic, you can use factoring or the quadratic formula to find the possible values of your trigonometric function.
- Range of Trigonometric Functions: Always remember the range of sine and cosine functions is $[-1, 1]$. Any potential solutions outside this range are invalid.
- Unit Circle is Crucial: Knowing your unit circle values and understanding which quadrants yield positive or negative trigonometric values is essential for finding all the solutions within a given interval.
Keep practicing, and you'll be solving these complex trig equations like a pro in no time! Let us know in the comments if you have any other tricky equations you'd like us to break down.
Understanding the Problem: A Deeper Dive
Alright, let's really unpack what we just did, guys. The equation $2 \sin^2(\theta) + 7 \cos(\theta) - 5 = 0$ might look intimidating at first glance because it mixes two different trigonometric functions, sine and cosine. The fundamental principle we employed is the reduction to a single trigonometric function. This is a cornerstone technique in solving trigonometric equations that involve multiple types of trig functions. The Pythagorean identity, $\sin^2(\theta) + \cos^2(\theta) = 1$, is the magic wand that allows us to perform this reduction. By expressing $\sin^2(\theta)$ as $1 - \cos^2(\theta)$, we effectively converted the entire equation into an expression solely dependent on $\cos(\theta)$. This transformation is critical because it allows us to treat the equation as a polynomial, specifically a quadratic, in terms of $\cos(\theta)$.
The process of substituting $x$ for $\cos(\theta)$ is a standard algebraic maneuver to simplify complex expressions and make them easier to visualize and solve. The resulting quadratic equation, $2x^2 - 7x + 3 = 0$, is then solved using familiar algebraic techniques. The factoring method we used, breaking down the middle term based on the product and sum of coefficients, is a testament to the underlying structure of quadratic equations. Successfully factoring this quadratic into $(2x - 1)(x - 3) = 0$ gave us two potential values for $x$, which are $x = 1/2$ and $x = 3$. These values are the potential solutions for $\cos(\theta)$.
The crucial step that follows is to translate these algebraic solutions back into the context of trigonometric functions and their properties. The equation $\cos(\theta) = 1/2$ has solutions because $1/2$ is within the valid range of the cosine function, which is $[-1, 1]$. To find these solutions within the specified interval $0 \ \textless \ \theta \ \textless \ 2\pi$, we rely on our knowledge of the unit circle and the behavior of the cosine function. Cosine represents the x-coordinate on the unit circle. We are looking for angles where the x-coordinate is $+1/2$. This occurs in the first and fourth quadrants. The principal value, or reference angle, for which $\cos(\theta) = 1/2$ is $\frac{\pi}{3}$ radians. In the first quadrant, the angle is directly the reference angle, so $\theta_1 = \frac{\pi}{3}$. In the fourth quadrant, the angle is $2\pi$ minus the reference angle, giving us $\theta_2 = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$. Both of these angles are within our specified domain.
Conversely, the equation $\cos(\theta) = 3$ has no solutions. This is because $3$ falls outside the range of the cosine function $[-1, 1]$. No matter what angle $\theta$ you choose, the value of $\cos(\theta)$ will never be $3$. This highlights the importance of checking the validity of your solutions against the inherent properties of the trigonometric functions you are working with. It’s a sanity check that prevents us from getting lost in the algebra and ensures our final answers are mathematically sound.
Finally, presenting the answers in radian measure is a standard convention in higher mathematics and calculus. It simplifies many formulas and theorems. Our final answers, $\frac{\pi}{3}$ and $\frac{5\pi}{3}$, are precisely in radians, fulfilling all the requirements of the problem. This systematic approach—simplification using identities, algebraic solution of the resulting polynomial, and back-substitution considering function properties and domain—is a powerful framework for tackling a wide variety of trigonometric equations.
Navigating the Unit Circle for Solutions
So, we've crunched the numbers and found that we're dealing with $\cos(\theta) = 1/2$ as our primary equation to solve for $\theta$ in the range $0 \ \textless \ \theta \ \textless \ 2\pi$. Now, this is where our trusty unit circle comes into play, guys. The unit circle is essentially a graphical representation of trigonometric functions where the radius is 1, centered at the origin (0,0) of a coordinate plane. For any point $(x, y)$ on the unit circle corresponding to an angle $\theta$ from the positive x-axis, we have the definitions: $\cos(\theta) = x$ and $\sin(\theta) = y$.
Our goal is to find the angles $\theta$ where the x-coordinate on the unit circle is $1/2$. We need to recall or look up the key angles on the unit circle where this happens. The cosine function is positive in the first and fourth quadrants. This is because in the first quadrant, both x and y coordinates are positive, and in the fourth quadrant, the x-coordinate is positive while the y-coordinate is negative. Since we're looking for $x = 1/2$, which is a positive value, we know our solutions must lie in these two quadrants.
The reference angle is the acute angle formed between the terminal side of the angle and the x-axis. For $\cos(\theta) = 1/2$, the reference angle is $\frac{\pi}{3}$ radians (or 60 degrees). This is a fundamental value that every math student should have memorized. It's the angle in the first quadrant where the cosine is $1/2$.
So, for the first quadrant, the angle $\theta$ is exactly equal to its reference angle. Therefore, our first solution is $\theta_1 = \frac{\pi}{3}$. This angle is between 0 and $2\pi$, so it's valid.
For the fourth quadrant, the angle is measured clockwise from the positive x-axis, or equivalently, it's an angle that is less than $2\pi$ but greater than $3\pi/2$ (the boundary between the third and fourth quadrants). To find the angle in the fourth quadrant that has the same cosine value as the reference angle, we subtract the reference angle from $2\pi$ (a full circle). This gives us: $\theta_2 = 2\pi - \textreference angle} = 2\pi - \frac{\pi}{3}$. To subtract these, we find a common denominator{3} - \frac{\pi}{3} = \frac{5\pi}{3}$. This angle, $\frac{5\pi}{3}$, is also within our interval $0 \ \textless \ \theta \ \textless \ 2\pi$.
We must also consider the case where $\cos(\theta) = 3$. As we've stressed, the cosine function's output is always confined to the interval $[-1, 1]$. Since $3$ is greater than $1$, there is no angle $\theta$ for which $\cos(\theta) = 3$. This is an important check; we don't want to waste time searching for solutions that simply don't exist. It's like trying to find a unicorn – they're not in the realm of possibility!
Therefore, the only valid solutions within the given interval are $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$. These are our final answers, expressed in radians as requested. The unit circle provides a clear and intuitive way to visualize and find these solutions once we've reduced the original equation to a simpler form like $\cos(\theta) = ext{constant}$. Keep practicing with your unit circle, and these types of problems will become much more manageable!