Advertisements
Advertisements
प्रश्न
Solve: 2cos2θ + sin θ - 2 = 0.
Advertisements
उत्तर
2cos2θ + sin θ - 2 = 0
⇒ 2( 1 - sin2θ) + sin θ - 2 = 0
⇒ 2 - 2 sin2θ + sin θ - 2 = 0
⇒ - sin θ( 2 sin θ - 1) = 0
⇒ sin θ( 2 sin θ - 1) = 0
⇒ sin θ = 0 or 2 sin θ - 1 = 0
⇒ sin θ = 0 or sin θ = `1/2`
⇒ θ = 30°
संबंधित प्रश्न
Without using trigonometric tables, evaluate the following:
`( i)\frac{\cos37^\text{o}}{\sin53^\text{o}}\text{ }(ii)\frac{\sin41^\text{o}}{\cos 49^\text{o}}(iii)\frac{\sin30^\text{o}17'}{\cos59^\text{o}\43'}`
Without using trigonometric tables evaluate the following:
`(i) sin^2 25º + sin^2 65º `
If tan 2θ = cot (θ + 6º), where 2θ and θ + 6º are acute angles, find the value of θ
Solve.
`cos22/sin68`
solve.
cos240° + cos250°
Evaluate.
`(sin77^@/cos13^@)^2+(cos77^@/sin13^@)-2cos^2 45^@`
Use tables to find sine of 21°
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
If tanθ = 2, find the values of other trigonometric ratios.
