Sin^2x+3cosx=0
To solve the equation \( \sin^2 x + 3 \cos x = 0 \), we can rewrite \( \sin^2 x \) in terms of \( \cos x \) using the Pythagorean identity:
\[ \sin^2 x = 1 - \cos^2 x \]
Substituting this into the equation gives us:
\[ 1 - \cos^2 x + 3 \cos x = 0 \]
Rearranging, we have:
\[ -\cos^2 x + 3 \cos x + 1 = 0 \]
Multiply the entire equation by -1 to make the leading coefficient positive:
\[ \cos^2 x - 3 \cos x - 1 = 0 \]
Next, we can treat this as a quadratic equation in terms of \( \cos x \). Using the quadratic formula, \( \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -3, c = -1 \):
\[ \cos x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ \cos x = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2} \]
This gives us two potential solutions for \( \cos x \):
\[ \cos x_1 = \frac{3 + \sqrt{13}}{2}, \quad \cos x_2 = \frac{3 - \sqrt{13}}{2} \]
Now we need to check if these values are within the valid range for \( \cos x \) (i.e., between -1 and 1).
For \( \cos x_1 = \frac{3 + \sqrt{13}}{2} \):
- Since \( \sqrt{13} \approx 3.605 \):
- Thus, \( 3 + \sqrt{13} \approx 6.605 \)
- Which gives \( \frac{6.605}{2} \approx 3.3025 \), which is not within the range \([-1, 1]\).
For \( \cos x_2 = \frac{3 - \sqrt{13}}{2} \):
- Again, calculating gives \( 3 - \sqrt{13} \approx -0.605 \)
- Thus, \( \frac{-0.605}{2} \approx -0.3025 \), which is within the range \([-1, 1]\).
Now we can find the angle \( x \) from \( \cos x = \frac{3 - \sqrt{13}}{2} \):
Let \( k \in \mathbb{Z} \) be an integer for the general solutions:
\[ x = \pm \cos^{-1}\left(\frac{3 - \sqrt{13}}{2}\right) + 2k\pi \]
So the complete set of solutions is:
\[ x = \pm \cos^{-1}\left(\frac{3 - \sqrt{13}}{2}\right) + 2k\pi, \quad k \in \mathbb{Z} \]
You can find the numerical value of \( \cos^{-1} \left(\frac{3 - \sqrt{13}}{2}\right) \) using a calculator for approximate angles if necessary.
