HELP! AP Calculus questions that I can't figure out

Solve for the indicated variable

2. cos2x+3cosx+2=0 ; x

$\small{\text{$ \begin{array}{rcl} \cos^2{(x)} +3 \cos{(x)} +2 &=& 0 \qquad | \qquad \cos{(x)} = z \qquad x = \pm\arccos{(z)}\\ z^2 +3z+2 &=& 0 \\ z_{1,2} &=& \dfrac{-3\pm\sqrt{9-4\cdot 2} }{2} \\ z_{1,2} &=& \dfrac{-3\pm 1 }{2} \\ z_{1} &=& -1 \\ z_{2} &=& -2 \qquad \text{ no solution} \\\\ x_{1,2} &=& \pm \arccos{(z_1)} \\ x_{1,2} &=& \pm\arccos{(-1)} \\ x_1 &=& \pi \\ x_2 &=& -\pi \\ \mathbf{x} & \mathbf{=} & \mathbf{\pi \pm 2k\pi } \qquad k \in N\\ \end{array} $}}$

Factor as indicated

1.  $(2x+1)^{\frac32}x^{\frac12}+(2x+1)^{\frac52}x^{-\frac{1}{2} } = (2x+1)^{\frac32}x^{-\frac{1}{2}}(........)$

$\small{ \begin{array}{rcl} (2x+1)^{\frac32}x^{\frac12}+(2x+1)^{\frac52}x^{-\frac{1}{2} } \\ &=& (2x+1)^{\frac32}x^{\frac{2-1}{2} } + (2x+1)^{\frac{2+3}{2}}x^{-\frac{1}{2} }\\ &=& (2x+1)^{\frac32}x^{\frac22}x^{-\frac{1}{2} } + (2x+1)^{\frac22} (2x+1)^{\frac{3}{2}}x^{-\frac{1}{2} }\\ &=& (2x+1)^{\frac32}x\cdot x^{-\frac{1}{2} } + (2x+1)\cdot (2x+1)^{\frac{3}{2}}x^{-\frac{1}{2} }\\ &=& (2x+1)^{\frac32} x^{-\frac{1}{2} } \left[ x + (2x+1) \right] \\ \mathbf{ (2x+1)^{\frac32}x^{\frac12}+(2x+1)^{\frac52}x^{-\frac{1}{2} } } &\mathbf{=} & \mathbf{(2x+1)^{\frac32} x^{-\frac{1}{2} } \left( 3x+1\right) }\\ \end{array} }$

Simplify each expression

1. (x1/2-x1/3) / x1/6

$\small{ \begin{array}{rcl} \dfrac{ x^\frac12-x^\frac13 } { x^\frac16 } \\ &=& \dfrac{ x^\frac12 } { x^\frac16 } - \dfrac{ x^\frac13 } { x^\frac16 }\\ &=& x^\frac12 x^{-\frac16 } - x^\frac13 x^{-\frac16} \\ &=& x^\frac36 x^{-\frac16 } - x^\frac13 x^{-\frac16} \\ &=& x^\frac26 - x^\frac13 x^{-\frac16} \\ &=& x^\frac13 - x^\frac13 x^{-\frac16} \\ \dfrac{ x^\frac12-x^\frac13 } { x^\frac16 } &=& x^\frac13 ( 1 - x^{-\frac16} )\\ \end{array} }$

2. (1-(sinx+cosx)2) / 2sinx

$\small{\text{$ \begin{array}{rcl} \dfrac{ 1-[\sin{(x)}+\cos{(x)}]^2 } { 2\sin{(x)} } \\ &=& \dfrac{ 1-[\sin^2{(x)}+2\cdot \sin{(x)} \cos{(x)} +\cos^2{(x)}] } { 2\sin{(x)} } \qquad | \qquad \sin^2{(x)}+ \cos^2{(x)} = 1\\ &=& \dfrac{ 1-[1+2\cdot \sin{(x)} \cos{(x)} ] } { 2\sin{(x)} } \\ &=& \dfrac{ 1-1-2\cdot \sin{(x)} \cos{(x)} } { 2\sin{(x)} } \\ &=& \dfrac{ -2\cdot \sin{(x)} \cos{(x)} } { 2\sin{(x)} } \\ \mathbf{\dfrac{ 1-[\sin{(x)}+\cos{(x)}]^2 } { 2\sin{(x)} } } & \mathbf{=}& \mathbf{-\cos{(x)} } \\ \end{array} $}}$

Factor completely

1. 5cos2x-5sin2x+cosx+sinx

$\small{\text{$ \begin{array}{rcl} & & 5\cos^2{(x)}-5\sin^2{(x)}+\cos{(x)}+\sin{(x)} \\ &=& 5 [ \cos^2{(x)}-\sin^2{(x)} ] + [ \cos{(x)}+\sin{(x)} ] \\ &=& 5 [ \cos{(x)}-\sin{(x)} ] [ \cos{(x)}+\sin{(x)} ]+ [ \cos{(x)}+\sin{(x)} ] \\ \mathbf{ 5\cos^2{(x)}-5\sin^2{(x)}+\cos{(x)}+\sin{(x)} }& \mathbf{=} & \mathbf{ [ \cos{(x)}+\sin{(x)} ] [ 5 ( \cos{(x)}-\sin{(x)} ) +1 ] }\\ \end{array} $}}$

Solve for the indicated variable

1. x4-4x2+2=0 ; 0

$\small{\text{$ \begin{array}{rcl} x^4-4x^2 +2 &=& 0 \qquad | \qquad x^2 = z \qquad x = \pm\sqrt{z}\\ z^2 -4z+2 &=& 0 \\ z_{1,2} &=& \dfrac{4\pm\sqrt{16-4\cdot 2} }{2} \\ z_{1,2} &=& \dfrac{4\pm\sqrt{8} }{2} \qquad | \qquad \sqrt{8} = \sqrt{4\cdot 2} = 2\sqrt{2}\\ z_{1,2} &=& \dfrac{4\pm2\sqrt{2} }{2} \\ z_{1,2} &=& 2 \pm\sqrt{2} \\ z_1 &=& 2 +\sqrt{2} \\ z_2&=& 2 -\sqrt{2} \\\\ x_{1,2} &=& \pm\sqrt{z_1} \\ x_{1,2} &=& \pm\sqrt{2 +\sqrt{2}}\\ \mathbf{ x_1 }&\mathbf{=}& \mathbf{\sqrt{2 +\sqrt{2}}}\\ \mathbf{x_2 }&\mathbf{=}& \mathbf{-\sqrt{2 +\sqrt{2}}}\\\\ x_{3,4} &=& \pm\sqrt{z_2} \\ x_{3,4} &=& \pm\sqrt{2 -\sqrt{2}}\\ \mathbf{x_3} &\mathbf{=}& \mathbf{\sqrt{2 -\sqrt{2}}}\\ \mathbf{x_4} &\mathbf{=}& \mathbf{-\sqrt{2 -\sqrt{2}}}\\\\ \end{array} $}}$