Advertisements
Advertisements
प्रश्न
Find the magnitude of angle A, if 2 cos2 A - 3 cos A + 1 = 0
उत्तर
2 cos2 A – 3 cos A + 1 = 0
2 cos2 A – cos A - 2cosA +1 = 0
cos A(2cos A – 1) – (2 cos A – 1) = 0
( 2 cosA – 1) (cos A – 1) = 0
2 cos A – 1 = 0 aand cos A – 1 = 0
cos A = `(1)/(2)` and cos A = 1
A = 60° and A = 0°
APPEARS IN
संबंधित प्रश्न
If sin 3A = 1 and 0 < A < 90°, find sin A
Solve the following equation for A, if sec 2A = 2
Solve for 'θ': `sin θ/(3)` = 1
If θ < 90°, find the value of: sin2θ + cos2θ
If `sqrt(3)` sec 2θ = 2 and θ< 90°, find the value of
cos2 (30° + θ) + sin2 (45° - θ)
In the given figure, if tan θ = `(5)/(13), tan α = (3)/(5)` and RS = 12m, find the value of 'h'.
Evaluate the following: `(sin62°)/(cos28°)`
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: cos72° - cos88°
Evaluate the following: tan(78° + θ) + cosec(42° + θ) - cot(12° - θ) - sec(48° - θ)
Prove the following: sin58° sec32° + cos58° cosec32° = 2
