How do you solve the following equation $6 {cos}^{2} x = 19 sin x + 16$ $6cos^2x = 19sinx + 16$ in the interval [0, 2pi]?

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How do you solve the following equation $6 {cos}^{2} x = 19 sin x + 16$ $6cos^2x = 19sinx + 16$ in the interval [0, 2pi]?

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Answer 1 · 2016-07-03T09:58:35

$x=(180^o)+41^o49'=221^o49' in QIII$
$Q IV, x=(360^o)-(41^o49')=318^o11'$ ,

Explanation:

$6cos^2x=19sinx+16$
$rArr 6(1-sin^2x)=19sinx+16$
$rArr 6sin^2x+19sinx+10=0$
$rArr 6sin^2x+15sinx+4sinx+10=0$
$rArr 3sinx(2sinx+5)+2(2sinx+5)=0$
$rArr (2sinx+5)(3sinx+2)=0$
$rArr sinx =-5/2,$ impossible , as, $sinx in [-1,1], sinx=-2/3$
$sinx =-2/3 ~= -0.6667 = -sin(41^o49')$

Since $sinx<0$ and, $x in [0,2pi]$ , x lies in Quadrant III, IV.

Accordingly, $x=(180^o)+41^o49'=221^o49' in QIII$
For $Q IV, x=(360^o)-(41^o49')=318^o11'$ ,

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How do you solve the following equation $6 {cos}^{2} x = 19 sin x + 16$ $6cos^2x = 19sinx + 16$ in the interval [0, 2pi]?

1 Answer

Explanation:

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How do you solve the following equation $6 {cos}^{2} x = 19 sin x + 16$ $6cos^2x = 19sinx + 16$ in the interval [0, 2pi]?