Question

if 2 sin theta=2- cos theta

Answer 1

2sin(θ) = 2 - cos(θ)——(i)

since, (sin(θ))^2 + (cos(θ))^2 = 1

=> (cos(θ))^2 = 1 - (sin(θ))^2

=> cos(θ) = square_root( 1 - (sin(θ))^2 )

therefore, equation (i) becomes

2sin(θ) = 2 - square_root( 1 - (sin(θ))^2 )

=> 2 - 2sin(θ) = square_root( 1 - (sin(θ))^2 )

squaring both sides , we get

( 2 - 2sin(θ))^2 = 1 - (sin(θ))^2

=> 4 - 8sin(θ) + 4(sin(θ))^2 = 1 - (sin(θ))^2

=> 5(sin(θ))^2 - 8sin(θ) + 3 = 0

which is a quadratic equation on sin(θ)

so using shreedhar Acharya’s formula for quadratic equations

ax^2 + bx +c =0 ==> x = [- b +_ square_root (b^2 - 4ac)]/2a, we get

sin(θ) = [-(-8) +_ square_root (8^2 -4*5*3)]/ 2*5

=> sin(θ) = [ 8 +_ square_root (64 - 60)] / 10

=> sin(θ) = [ 8 +_ square_root (4)] / 10

=> sin(θ) = [ 8 +_ 2] / 10

therefore, sin(θ) = (8+2)/10 , (8–2) / 10

or sin(θ) = 1 , 0.6

Answer 2

Answer:

Step-by-step explanation:

2sin(θ) = 2 - cos(θ)——(i)

since, (sin(θ))^2 + (cos(θ))^2 = 1

=> (cos(θ))^2 = 1 - (sin(θ))^2

=> cos(θ) = square_root( 1 - (sin(θ))^2 )

therefore, equation (i) becomes

2sin(θ) = 2 - square_root( 1 - (sin(θ))^2 )

=> 2 - 2sin(θ) = square_root( 1 - (sin(θ))^2 )

squaring both sides , we get

( 2 - 2sin(θ))^2 = 1 - (sin(θ))^2

=> 4 - 8sin(θ) + 4(sin(θ))^2 = 1 - (sin(θ))^2

=> 5(sin(θ))^2 - 8sin(θ) + 3 = 0

which is a quadratic equation on sin(θ)

so using shreedhar Acharya’s formula for quadratic equations

ax^2 + bx +c =0 ==> x = [- b +_ square_root (b^2 - 4ac)]/2a, we get

sin(θ) = [-(-8) +_ square_root (8^2 -4*5*3)]/ 2*5

=> sin(θ) = [ 8 +_ square_root (64 - 60)] / 10

=> sin(θ) = [ 8 +_ square_root (4)] / 10

=> sin(θ) = [ 8 +_ 2] / 10

therefore, sin(θ) = (8+2)/10 , (8–2) / 10

or sin(θ) = 1 , 0.6

Thx