Please Help!!!! Find all solutions in the interval [0, 2π).? 1. 7 tan^3(x) - 21 tan x = 0 2. 2 sin^2(x) = sin x 3. Find all solutions to the equation. Please show your work. cos2x + 2 cos x + 1 = 0
@rsnoob can u help?
@laura*
@sourwing can you please help me figure this out
Problem 1 use algebra to rearrange and make into a simpler function. Then graph the function and find the x-intercepts. \[7\tan ^{3}x-21tanx=0\] \[7\tan ^{3}x=21tanx\] \[\tan ^{3}x=3tanx\] \[\frac{ \tan ^{3}x }{ 3tanx }=0\] \[\frac{ \tan ^{2}x }{ 3 }=0\] \[\tan ^{2}x=0\]
You can do similar elimination of terms with the remaining equations.. For question 1 remember that the only time that tan^2 can be zero is then tanx is zero. Therefore your answer will be all places where tanx=0 on your interval (This should be a memorized graph for most Algebra II/Pre-Calc/Calc I classes).
but these are my choices: pie/3 , 2pi/3, 4pi/3, 5pi/3 0, pi/5, π, 6pi/5 0, pi/3, 2pi/3, π, 4pi/3, 5pi/3 0, pi/3, π, 4pi/3
The Original Function Graph is attached. So evaluating by looking at the graph would be the 3rd response. (I used google to get the graph)
thankyou thankyou!! can you help with the others???
I thought it was that but i wasn't quite sure
2 * sin^2(x) = sin(x) dividing both sides of equation by sin(x) gets: (2* sin^2(x))/sin(x) = 1 dividing both sides of eqution by 2 gets: sin^2(x)/sin(x) = 1/2 since sin^2(x) = sin(x) * sin(x), equation becomes: (sin(x)*sin(x))/sin(x) = 1/2 which becomes: sin(x) = 1/2 ----- if sin(x) = 1/2, then x = 30 degrees. ----- since sin(x) = hypotenuse / y value on the graph, and since hypotenuse is always positive, and since y value on the graph is positive in quadrants 1 and 2, then sin (180-x) = sin(x). ----- since 180-30 = 150, then x can be either 30 degrees or 150 degrees. Ooops I forgot you were doing radians so just convert
I will help with the third if you rewrite using the double angle identity for cos and try some of the elimination steps I have shown.
I'm at cos^2x=-2cos-1
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