Question

Cos²x+2=sin²x+3cosx​

Answer 1

$\cos^{2} x +2= { \sin}^{2} x + 3 x\cos x \\ \cos^{2} x + 2 = 1 - \cos^{2} x + 3\cos x \\ 2cos^{2} x - 3 \cos x + 1 = 0 \\ ( \cos x - 1)(2 \cos x - 1) = 0 \\ \cos x = 1 \: or \: \cos x = \frac{1}{2}$

Answer 2

To solve the equation $\cos^2(x) + 2 = \sin^2(x) + 3\cos(x)$, we can use trigonometric identities to simplify and solve for $x$.

1. Start by applying the Pythagorean trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ to the equation:

   $\cos^2(x) + 2 = 1 + 3\cos(x)$

2. Subtract 1 from both sides of the equation to isolate the terms involving $\cos(x)$:

   $\cos^2(x) + 2 - 1 = 3\cos(x)$

3. Simplify:

   $\cos^2(x) + 1 = 3\cos(x)$

4. Now, we have a quadratic equation in terms of $\cos(x)$. To solve for $\cos(x)$, move all terms to one side of the equation:

   $\cos^2(x) - 3\cos(x) + 1 = 0$

5. You can solve this quadratic equation using the quadratic formula:

   $\cos(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   Here, $a = 1$, $b = -3$, and $c = 1$. Plug these values into the formula:

   $\cos(x) = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$

   $\cos(x) = \frac{3 \pm \sqrt{9 - 4}}{2}$

   $\cos(x) = \frac{3 \pm \sqrt{5}}{2}$

6. So, you have two possible values for $\cos(x)$:

   $\cos(x) = \frac{3 + \sqrt{5}}{2}$ and $\cos(x) = \frac{3 - \sqrt{5}}{2}$

7. Now, to find the corresponding values of $x$, you can use the inverse cosine function:

   x = \arccos\left(\frac{3 + \sqrt{5}}{2} and x = \arccos\left(\frac{3 - \sqrt{5}}{2}

These are the solutions for $x$. Note that the solutions may involve multiple values of $x$ due to the periodic nature of trigonometric functions.