trig equation (1 Viewer)

philly2000 · Apr 28, 2009

cos(4x) - 7sin^2 (2x) = 0 solve for 0 =< x =< 360

thanks

Drongoski · Apr 28, 2009

cos 4x - 7 sin²2 x = 0

.: (1 - 2sin²2x) - 7 sin²x = 0

i.e. 1 - 9 sin²2x = 0

i.e. (1-3sin 2x)(1 + 3sin 2x) = 0

.: either (1): 1 - 3sin 2x = 0 ==> sin 2x = 1/3

or (2): 1 + sin 2x = 0 ==> sin 2x = -1/3

For (1): 2x = 19.5, 160.5, 369.5, 520.5 deg.

For (2): 2x = 199.5, 340.5, 559.5, 700.5 deg

For x, just divide by 2

study-freak · Apr 28, 2009

$\\cos4x-7sin^22x=0 \\1-9sin^22x=0 \\sin^22x=\frac{1}{9} \\sin2x=\pm \frac{1}{3} \\2x=180^\circ n\pm sin^{-1}\frac{1}{3} \\x=90^\circ n\pm \frac{sin^{-1}\frac{1}{3}}{2}\approx$ 9^\circ44', 80^\circ16', 99^\circ44', 170^\circ16', 189^\circ44', 260^\circ16', 279^\circ44', 350^\circ16' \\$ for 0^\circ \leq x\leq 360^\circ$
For a nice looking solution

philly2000 · Apr 28, 2009

gah thanks guys, cant believe i couldnt see that >->'

trig equation (1 Viewer)

philly2000

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Drongoski

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study-freak

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philly2000

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