Question

Use trigonometric identities to solve the equation 2sin(2θ)-2cos(θ)=0 exactly for 0≤θ≤2π.

A.) What is 2sin(2θ) in terms of sin(θ)and cos(θ)?

B.) After making the substitution from part 1, what is the common factor for the left side of the expression 2sin(2θ)-2cos(θ)=0 ?

C.) Choose the correctly factored expression from below.

a.) $cos(\Theta )(4sin(\Theta )-2)=0$

b.) $4 sin (\Theta ) cos(\Theta ) = 2 cos (\Theta )$

c.) $2 cos (\Theta ) 2 sin (\Theta ) = 0$

d.) $2cos(\Theta )(2sin(\Theta )-1) = 0$

Answer 1

Answer: Given equation 2 Sin (20)-20050 =0 a) given that 2 sin (20) in terms of sin(o) and cosco) According to formula sin(2A) = 2 Sin(A) COB(A) 2 sin(20) = 2 sin (20) = 2 (2 sino coso] 2 fin (20) = 4 Sinco) COS(O) 2 Sin (20)= 4 Sinco) cos (O) b) Given Expression 2 Sin (20)-2 Cos (0) 50 substituting 2 sin (20) = 4 Sin() coco) from part 1 in above Expression

2 Sin (20)-2 COJCO)=0 4 sinco) Cos (0)-2 Cos (0) 50 2 cos (o) [2 Sin (0)-1] =0 common factor for the left side of the expression = 2 Cosco) c> from option (a) we take coso (4 Sonco) - 2) =0 4 8inco Cos (0)-2 CO3(0)50 (2)x2 8inco) Cerce) – 2 Cosco) so but 2 fonco) cosco) s fin (20) 2 (sin(20) – 2 Cos@so 2857 (20)-2 COCO) 50 He already given 28in (20) -2 Cos (0) 50 thats why option (Q) is correctly factored expression