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. \( 2 \cos ^{2} t+3 \cos t+1=0 \)
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For the first equation, \( 2 \sin x + \sqrt{3} = 0 \), you can rearrange it to find \( \sin x = -\frac{\sqrt{3}}{2} \). The solutions in the interval \( [0, 2\pi] \) occur where the sine function is negative, specifically in the third and fourth quadrants, giving us \( x = \frac{4\pi}{3} \) and \( x = \frac{5\pi}{3} \). For the second equation, \( (\cos x - 1)(\sin x + 1) = 0 \), you can separate it into two parts. Setting \( \cos x - 1 = 0 \) leads to \( \cos x = 1 \), giving \( x = 0 \). Setting \( \sin x + 1 = 0 \) leads to \( \sin x = -1 \), giving \( x = \frac{3\pi}{2} \). Thus, the solutions in \( [0, 2\pi] \) are \( x = 0 \) and \( x = \frac{3\pi}{2} \). For the third equation \( 2 \cos^2 t + 3 \cos t + 1 = 0 \), this is a quadratic in terms of \( \cos t \). Letting \( u = \cos t \), the equation becomes \( 2u^2 + 3u + 1 = 0 \). Using the quadratic formula, you find \( u = \frac{-3 \pm \sqrt{3}}{4} \). This gives two values for \( \cos t \), which you can further analyze to find the corresponding angles that satisfy \( t \) between \( 0 \) and \( 2\pi \).