Question

2sin^2x- cosx-1 =0 [3pi;4pi]

Answer 1

2-2cos²x-cosx-1=0
2cos²x+cosx-1=0
cosx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒cosx=-1⇒x=π+2πn
a2=(-1+3)/4=1/2⇒x=+-π/3+2πn
x=π U x=23π/6

Answer 2

To solve the equation $2\sin^2(x) - \cos(x) - 1 = 0$ in the interval $[3\pi, 4\pi]$, you can use trigonometric identities and techniques. Here's how to do it step by step:

1. First, let's rewrite the equation using trigonometric identities. Recall that $\sin^2(x) = 1 - \cos^2(x)$. So, we can rewrite the equation as:

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

2. Now, simplify the equation:

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

3. Combine like terms:

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

4. Next, let's make a substitution to simplify this quadratic equation. Let $u = \cos(x)$. Now, the equation becomes:

   $-2u^2 - u + 1 = 0$

5. Solve the quadratic equation for $u$:

   To solve this quadratic equation, you can use the quadratic formula:

   $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   Here, $a = -2$, $b = -1$, and $c = 1$. Plug in these values:

   $u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(1)}}{2(-2)}$

   $u = \frac{1 \pm \sqrt{1 + 8}}{-4}$

   $u = \frac{1 \pm \sqrt{9}}{-4}$

   $u = \frac{1 \pm 3}{-4}$

   So, you have two possible values for $u$:

   $u_1 = \frac{1 + 3}{-4} = -1$

   $u_2 = \frac{1 - 3}{-4} = \frac{1}{2}$

6. Now, recall that $u = \cos(x)$. So, you have two possible values for $\cos(x)$:

   $1. \cos(x) = -1$ $2. \cos(x) = \frac{1}{2}$

7. Solve for $x$ in each case:

   7.1. For $\cos(x) = -1$, you are looking for angles $x$ where the cosine is -1. In the interval $[3\pi, 4\pi]$, the cosine is equal to -1 at $x = 3\pi$.

   7.2. For $\cos(x) = \frac{1}{2}$, you are looking for angles $x$ where the cosine is $\frac{1}{2}$. In the interval $[3\pi, 4\pi]$, the cosine is equal to $\frac{1}{2}$ at $x = \frac{7\pi}{3}$ and $x = \frac{11\pi}{3}$.

So, the solutions to the equation $2\sin^2(x) - \cos(x) - 1 = 0$ in the interval $[3\pi, 4\pi]$ are $x = 3\pi$, $x = \frac{7\pi}{3}$