Which of the following is NOT a valid strategy for solving the given equation?a. To solved the equation cos²θ= sinθcosθ, first subtract sinθcosθ from both sides, then factor out the common factor of cosθ on the left side.b. To solve the equation cos²θ=sinθcosθ, ***FIRST DIVIDE**** both sides by cosθ.c. To solve sin2θ+2cosθsin2θ=0, first factor out the common factor of sin2θ on the left side.d. To solve the equation sinθ+cosθ=1, first square both sides of the equation.

Answer:The correct choice is B.Step-by-step explanation:a. Given [tex]\cos^2(\theta)=\sin(\theta)\cos(\theta)[/tex]We can subtract [tex]\sin(\theta)\cos(\theta)[/tex] from both sides to obtain;[tex]\cos^2(\theta)-\sin(\theta)\cos(\theta)=0[/tex]We can then factor [tex]\cos(\theta)[/tex] to obtain;[tex]\cos(\theta)(\cos(\theta)-\sin(\theta))=0[/tex]We can then proceed with our solution using the zero product principle.b. Given [tex]\cos^2(\theta)=\sin(\theta)\cos(\theta)[/tex], it is not valid to divide both sides by [tex]\cos(\theta)[/tex] because [tex]\cos(\theta)[/tex] could be equal to zero and division by zero in disguise will not result in the true solutions of the given equation.c. To solve sin2θ+2cosθsin2θ=0, first factor out the common factor of sin2θ on the left side. This is also a justifiable approach.d. To solve the equation sinθ+cosθ=1, first square both sides of the equation.Squaring both sides will help solve the equation with double angle properties easily.