Question

If α is a root of 25cos^2θ + 5cosθ -12 = 0, π/2 < α < π, then sin2α is equal to *

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: given \: trigo \: quadratic \: equation \: is \: 25cos^2θ + 5cosθ -12 = 0$

$also \: α \: is \: a \: root of \: 25cos^2θ + 5cosθ -12 = 0$

$but \: here \: \: theta \: is \: its \: permissible \: and \: posible \: given \: root$

$thus \: then \: \alpha = theta$

$so \: now \: thus \: considering \: 25cos^2θ + 5cosθ -12 = 0$

$= 25 \: \cos \: square \: theta + 20 \: cos \: theta - 15 \: cos \: theta - 12 = 0$

$= 5cos \: theta(5cos \: theta + 4) -3(5cos \: theta + 4) = 0$

$= (5cos \: theta + 4)(5cos \: theta - 3) = 0$

$ie \: cos \: theta = - 4 \div 5 \\ or \\ cos \: theta =3 \div 5$

$but \: as \: π/2 < α < π \\ alpha \: ie \: theta \: lies \: in \: quadrant \: 2$

$so \: in \: quadrant \: 2 \\ sine \: and \: cosec \: are \: positive \: \: and \: other \: trigonometric \: ratios \: are \: negative$

$thus \: cos \: \alpha = 3 \div 5 \: is \: absurd \\ so \: cos \: \alpha = - 4 \div 5$

$thus \: then \: using \\ sin \: square \: \alpha + cos \: square \: \alpha = 1$

$so \: sin \: square \: \alpha = 1 - cos \: square \: \alpha$

$substitute \: value \: of \: cos \: \alpha \\ we \: get \: sin \: square \: \alpha = 3 \div 5$

$hence \: value \: of \: sin \: square \: alpha \: is \: 3 \div 5$