# Solve (2sinxcotx+sinx+4cotx-2)/(2cotx+1)=sinx-2

Solve $\left(2\mathrm{sin}x\mathrm{cot}x+\mathrm{sin}x+4\mathrm{cot}x-2\right)/\left(2\mathrm{cot}x+1\right)=\mathrm{sin}x-2$

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Step 1 Hence, given equation is true.

Step 2

Factors of $2\mathrm{sin}\left(x\right)\mathrm{cot}\left(x\right)+\mathrm{sin}\left(x\right)-4\mathrm{cot}\left(x\right)-2is=2\mathrm{cot}\left(x\right)\mathrm{sin}\left(x\right)-2\cdot 2\mathrm{cot}\left(x\right)-2=2\mathrm{cot}\left(x\right)\left(\mathrm{sin}\left(x\right)-2\right)+\mathrm{sin}\left(x\right)-2=2\left(-2+\mathrm{sin}\left(x\right)\right)\mathrm{cot}\left(x\right)+1\cdot \left(-2+\mathrm{sin}\left(x\right)\right)=\left(-2+\mathrm{sin}\left(x\right)\right)\left(2\mathrm{cot}\left(x\right)+1\right)\left(2\mathrm{sin}\left(x\right)\mathrm{cot}\left(x\right)+\mathrm{sin}\left(x\right)-4\mathrm{cot}\left(x\right)-2\right)/\left(2\mathrm{cot}\left(x\right)+1\right)=-2+\mathrm{sin}\left(x\right)$