Find all solution of the equations in the interval \( [0,2 \pi] \) 11. \( 2 \sin x+\sqrt{3}=0 \) 12. \( (\cos x-1)(\sin x+1)=0 \) 13. \( 2 \cos ^{2} t+3 \cos t+1=0 \)

Let's solve each equation one by one in the interval \( [0, 2\pi] \). ### 11. \( 2 \sin x + \sqrt{3} = 0 \) First, we isolate \( \sin x \): \[ 2 \sin x = -\sqrt{3} \] \[ \sin x = -\frac{\sqrt{3}}{2} \] The sine function is negative in the third and fourth quadrants. The angles where \( \sin x = -\frac{\sqrt{3}}{2} \) are: \[ x = \frac{4\pi}{3}, \quad \frac{5\pi}{3} \] Thus, the solutions for this equation in the interval \( [0, 2\pi] \) are: \[ \boxed{\left\{ \frac{4\pi}{3}, \frac{5\pi}{3} \right\}} \] ### 12. \( (\cos x - 1)(\sin x + 1) = 0 \) This equation can be solved by setting each factor to zero. 1. **For \( \cos x - 1 = 0 \)**: \[ \cos x = 1 \] The solution in the interval \( [0, 2\pi] \) is: \[ x = 0 \] 2. **For \( \sin x + 1 = 0 \)**: \[ \sin x = -1 \] The solution in the interval \( [0, 2\pi] \) is: \[ x = \frac{3\pi}{2} \] Thus, the solutions for this equation in the interval \( [0, 2\pi] \) are: \[ \boxed{\{ 0, \frac{3\pi}{2} \}} \] ### 13. \( 2 \cos^2 t + 3 \cos t + 1 = 0 \) This is a quadratic equation in terms of \( \cos t \). Let \( u = \cos t \). The equation becomes: \[ 2u^2 + 3u + 1 = 0 \] We can use the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ u = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \] \[ u = \frac{-3 \pm \sqrt{9 - 8}}{4} \] \[ u = \frac{-3 \pm 1}{4} \] Calculating the two possible values: 1. \( u = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2} \) 2. \( u = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \) Now we find \( t \) for each value of \( u \): 1. **For \( \cos t = -\frac{1}{2} \)**: The angles in the interval \( [0, 2\pi] \) are: \[ t = \frac{2\pi}{3}, \quad \frac{4\pi}{3} \] 2. **For \( \cos t = -1 \)**: The angle in the interval \( [0, 2\pi] \) is: \[ t = \pi \] Thus, the solutions for this equation in the interval \( [0, 2\pi] \) are: \[ \boxed{\left\{ \frac{2\pi}{3}, \frac{4\pi}{3}, \pi \right\}} \]