Advertisements
Advertisements
Question
Solve: 2cos2θ + sin θ - 2 = 0.
Solution
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°
APPEARS IN
RELATED QUESTIONS
If tan 2θ = cot (θ + 6º), where 2θ and θ + 6º are acute angles, find the value of θ
if `tan theta = 3/4`, find the value of `(1 - cos theta)/(1 +cos theta)`
Use tables to find sine of 10° 20' + 20° 45'
Evaluate:
`(5sin66^@)/(cos24^@) - (2cot85^@)/(tan5^@)`
Evaluate:
`(3sin72^@)/(cos18^@) - sec32^@/(cosec58^@)`
Find the sine ratio of θ in standard position whose terminal arm passes through (3, 4)
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ −\[\sqrt{3} \tan 3\theta\] is equal to
Find the value of the following:
`((cos 47^circ)/(sin 43^circ))^2 + ((sin 72^circ)/(cos 18^circ))^2 - 2cos^2 45^circ`
Find the value of the following:
sin 21° 21′
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
