Solve the equation $\sin^2(\theta) + 5\cos(\theta) = 7$ for $\cos(\theta)$.
Understand the Problem
The question requires us to solve a trigonometric equation for $\cos(\theta)$. We are given $\sin^2(\theta) + 5\cos(\theta) = 7$. We can use the identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to rewrite the equation in terms of only $\cos(\theta)$. Then, we can solve the resulting quadratic equation.
Answer
No solutions exist for $\cos(\theta)$.
Answer for screen readers
No solutions exist for $\cos(\theta)$.
Steps to Solve
- Rewrite $\sin^2(\theta)$ using the Pythagorean identity
We know that $\sin^2(\theta) + \cos^2(\theta) = 1$, so $\sin^2(\theta) = 1 - \cos^2(\theta)$. Substitute this into the given equation:
$1 - \cos^2(\theta) + 5\cos(\theta) = 7$
- Rearrange the equation into a quadratic equation
Rearrange the equation to get a standard quadratic form:
$-\cos^2(\theta) + 5\cos(\theta) - 6 = 0$
Multiply by $-1$ to simplify:
$\cos^2(\theta) - 5\cos(\theta) + 6 = 0$
- Solve the quadratic equation for $\cos(\theta)$
Let $x = \cos(\theta)$. The equation becomes:
$x^2 - 5x + 6 = 0$
Factor the quadratic equation:
$(x - 2)(x - 3) = 0$
So, $x = 2$ or $x = 3$.
- Determine the valid solutions for $\cos(\theta)$
Since $x = \cos(\theta)$, we have $\cos(\theta) = 2$ or $\cos(\theta) = 3$. However, the range of the cosine function is $-1 \le \cos(\theta) \le 1$. Therefore, neither $\cos(\theta) = 2$ nor $\cos(\theta) = 3$ are valid solutions. Thus, there are no solutions for $\cos(\theta)$.
No solutions exist for $\cos(\theta)$.
More Information
The range of the cosine function, $\cos(\theta)$, is always between -1 and 1, inclusive. This means that $\cos(\theta)$ can never be equal to 2 or 3.
Tips
A common mistake is to correctly solve the quadratic equation but then fail to recognize that the resulting values for $\cos(\theta)$ are outside the valid range $[-1, 1]$. Always check if the solutions you obtain are within the possible range for the trigonometric functions.
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