Asked by LiNnda
8sin^2x+sinxcosx+cos^2x=4
Answers
Answered by
Reiny
very tricky
8sin^2x+sinxcosx+cos^2x=4
8sin^2x+sinxcosx+cos^2x=4(sin^2x + cos^2x)
8sin^2x+sinxcosx+cos^2x=4sin^2x + 4cos^2x
4sin^2x + sinxcosx - 3cos^2x = 0
(sinx + cosx)(4sinx - 3cosx) = 0
sinx = -cosx or sinx = (3/4)cosx
sinx/cosx = -1 or sinx/cosx = 3/4
tanx = -1 or tanx = 3/4
x = 135°, 315° or x = 36.9° or 216.9°
if you want radians, change your calculator to rad
8sin^2x+sinxcosx+cos^2x=4
8sin^2x+sinxcosx+cos^2x=4(sin^2x + cos^2x)
8sin^2x+sinxcosx+cos^2x=4sin^2x + 4cos^2x
4sin^2x + sinxcosx - 3cos^2x = 0
(sinx + cosx)(4sinx - 3cosx) = 0
sinx = -cosx or sinx = (3/4)cosx
sinx/cosx = -1 or sinx/cosx = 3/4
tanx = -1 or tanx = 3/4
x = 135°, 315° or x = 36.9° or 216.9°
if you want radians, change your calculator to rad
Answered by
LiNnda
thank u very much!
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