BQ :

2sin²Ө-cosӨ=1 )

The correct answers are 60, 180, 300

Where are you getting your incorrect answers.

sin^2(Ө) = 1- cos^2(Ө)

2*(1 - cos^2(Ө) - cos(Ө) = 1

2 - 2*cos^2(Ө) - cos(Ө) = 1

rearrange equation

2*cos^2(Ө) + cos(Ө) + 1 - 2 = 0

2*cos^2(Ө) + cos(Ө) - 1 = 0

use quadratic equation ,

let u = cos(Ө)

2u^2 + u -1 =0

u = (-1 +/- sqrt( 1 - 4*2*(-1) ) / 4

u= (-1 +/- sqrt ( 1 +8) ) /4

u = (-1 + /- 3 ) /4 =

u = (-4/4) or u = (-1 +3)/4 = 1/2

u =-1 or u = 1/2

cos(Ө) = -1

acos(1) = -pi

theta = -pi

cos(Ө) = 1/2

acos(1/2) = 60 and -60 degrees which is the same as 300

so the answer's are

60, 180, and 330

its 7/(5√2)

Not sure if this is allowed, but I am assuming that α and β are in quadrant I.

cosα = ⅘

sinα = ⅗

cosα = 7/(5√2) = 7√2/10

sinα = 1/(5√2) = √2/10

cos(α+β) = cosαcosβ - sinαsinβ

= (⅘)(7√2/10) - (⅗)(√2/10)

= 28√2/50 - 3√2/50

= √2/2

α+β = arccos(√2/2) = π/4