Question

If 2cos^2theta +sin theta-2=0 and 0°great than or equal to theta greater than or equal to 90°;find the value of theta

Answer 1

Answer:

θ = 30 °

Step-by-step explanation:

Heya friend !

Here it goes:

2 $cos^{2}$ θ + sin θ - 2 = 0 ,

where 0° ≥ θ ≤ 90°

Now,

2 (1 - $sin^{2}$) θ + sin θ - 2 = 0,

as, $sin^{2}$ θ + $cos^{2}$ θ = 1 . So,

    $cos^{2}$ θ = 1 - $sin^{2}$ θ

2 - 2 $sin^{2}$ θ + sin θ -2 = 0

sin θ - 2 $sin^{2}$ θ = 0

sin θ (1 - 2 sin θ) = 0                  

1 - 2 sin θ = 0

- 2 sin θ = - 1

So,

sin θ = 1 / 2

and,

we know,

sin 30° = 1/2

So,

θ = 30 °

Thanks,

Manav

Answer 2

Answer:

θ ≤ 90°

θ ≤ 90°Now,

θ ≤ 90°Now,2 (1 - sin^{2}sin

θ ≤ 90°Now,2 (1 - sin^{2}sin 2

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So,

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2 θ + sin θ -2 = 0

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2 θ + sin θ -2 = 0sin θ - 2 sin^{2}sin

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2 θ + sin θ -2 = 0sin θ - 2 sin^{2}sin 2

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2 θ + sin θ -2 = 0sin θ - 2 sin^{2}sin 2 θ = 0

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2 θ + sin θ -2 = 0sin θ - 2 sin^{2}sin 2 θ = 0sin θ (1 - 2 sin θ) = 0

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2 θ + sin θ -2 = 0sin θ - 2 sin^{2}sin 2 θ = 0sin θ (1 - 2 sin θ) = 0 1 - 2 sin θ = 0

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2 θ + sin θ -2 = 0sin θ - 2 sin^{2}sin 2 θ = 0sin θ (1 - 2 sin θ) = 0 1 - 2 sin θ = 0- 2 sin θ = - 1

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2 θ + sin θ -2 = 0sin θ - 2 sin^{2}sin 2 θ = 0sin θ (1 - 2 sin θ) = 0 1 - 2 sin θ = 0- 2 sin θ = - 1So,

θ ≤ 90°Now,2 (1 - sin^{2}sin 2 ) θ + sin θ - 2 = 0,as, sin^{2}sin 2 θ + cos^{2}cos 2 θ = 1 . So, cos^{2}cos 2 θ = 1 - sin^{2}sin 2 θ2 - 2 sin^{2}sin 2 θ + sin θ -2 = 0sin θ - 2 sin^{2}sin 2 θ = 0sin θ (1 - 2 sin θ) = 0 1 - 2 sin θ = 0- 2 sin θ = - 1So,sin θ = 1 / 2

sin 30° = 1/2