3. Solve tan(x−2π​)=3​1​ for all x∈R. tan(x−2π​)=3​1​ tan(x−3π​)=33​​ General solution: tan(x−3π​)=33​​ x−3π​x​=6π​+πn=2

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3. Solve tan(x−2π​)=3​1​ for all x∈R. tan(x−2π​)=3​1​ tan(x−3π​)=33​​ General solution: tan(x−3π​)=33​​ x−3π​x​=6π​+πn=2π​+πn​
4. Solve the equation sin(2x−0.35)=cos(3x) where 0≤x≤π. −0.35)=sin(2π​−3x)⋅cos(x)=sin(2π​−x) −0.35=2π​−3x+2πn and 2x−0.35=π−(2π​−3x)+2πn⇒0≤x
6. Solve cos(2θ)=sin(θ) algebraically where θ∈R. 1−2sin2θ=sin(θ)[cos(2θ)=1−2sin2θ] 7) 2sin2θ+sinθ−1=0 ⇒2sin2θ+2sinθ−sinθ−1=0 [severoting middue tern ⇒2sinθ(sinθ+1)−1(sinθ+1)=0 ⇒(2sinθ−1)(sinθ+1)=0 →sinθ=1/2, also can be, sinθ=−1
7. For what values of x is the expression 1−2cos2xtan(x)​→⎩⎨⎧​1−2cos2x=0cos2x=21​cosx=±21​​ \begin{tabular}{ll|l} If rositive: & & If negative: \\ cos(x)=2​1​ & ∴x∈[0,π] & cosx=2​−1​ \\ & x=4π​ & x=43π​∴⋅cos43π​ \end{tabular}
11. Prove the following identities: a. tan2x−sin2x=sin2xtan2x =tan2x(1−cos2x)(⋯sin2x+sin2x⋅cos2x=tan2x−tan2xsin2x=tan2x−xcos2x​(−⋅tanx=tan2x−sin2x∴tan2x−sin2x=tan2xsin2x​