$$\frac{\sin\left(6x\right)}{\sin\left(3x\right)}-\frac{\cos\left(6x\right)}{\cos\left(3x\right)}=\sec\left(3x\right)$$

$$f\left(x\right)=\frac{1}{5\left(2x^3+3x^2-12x\right)}$$

$$3+4x<x+2$$

$$4\left(1-z\right)^3$$

$$\int\cos^5bxdx$$

$$\frac{d}{dx}x^2y=15$$

$$4+-1+4+-2$$