Question

pls help me plz .....m........

Answer 1

6)

Starting with: 2sin(θ)=2−cos(θ)2sin(θ)=2−cos(θ)

Square both sides of the equation:

4sin2(θ)=4−4cos(θ)+cos2(θ)4sin2(θ)=4−4cos(θ)+cos2(θ)

As sin2(θ)=1−cos2(θ)sin2(θ)=1−cos2(θ), we can rewrite the equation as:

4−4cos2(θ)=4−4cos(θ)+cos2(θ)4−4cos2(θ)=4−4cos(θ)+cos2(θ)

Adding 4cos2(θ)−44cos2(θ)−4 to both sides of the equation:

0=5cos2(θ)−4cos(θ)0=5cos2(θ)−4cos(θ)

Factorising: cos(θ).[5cos(θ)−4]=0cos(θ).[5cos(θ)−4]=0

cos(θ)∈{0,0.8}cos(θ)∈{0,0.8}

cos2(θ)∈{0,0.64}cos2(θ)∈{0,0.64}

sin2(θ)∈{1,0.36}sin2(θ)∈{1,0.36}

sin(θ)∈{−1,1,−0.6,0.6}

Answer 2

hope this helps you☺️