how to solve -27(cos^2 3t)(sint) = 0 and 26(sin2t)(cos2t)=0?

Answer 1

If -27(cos^2 3t)(sin t) = 0, then either cos^2 3t or the sin t must equal zero.

sin t = 0 at either t=0 or t=pi

cos^2 3t = 0 when 3t = pi/2, 3pi/2, 5pi/2, 7pi/2, 9pi/2, 11pi/2 so
t = pi/6, pi/2, 5pi/6, 7pi/6, 3pi/2 and 11pi/6

So the enter answer is

t = 0, pi/6, pi/2, 5pi/6, pi, 7pi/6, 3pi/2 and 11pi/6

Solve the second part the exact same way.

Answer 2

One easy way to begin a problem where a product of terms gives zero, is to remember that the only way to get 0 is for one of the terms to be 0.

In the case of -27(cos^2 3t)(sin t)= 0

We know that -27 cannot be 0.

That leaves cos^2 3t = 0
(which is whenever cos 3t = 0, therefore 3t = (2k-1)pi/2

and
sin t = 0
t = k pi
(in degrees, t = -180, 0, 180, 360...)

(k = any integer)

Answer 3

-27(cos^2 3t)(sint) = 0 so
cos^2 3t = 0 or sin t = 0
cos^2 3t = 0 means that cos 3t = 0
Now cosx equals zero when x = pi/2 or 3pi/2
so cos 3t = 0 when 3t = pi/2 or 3t = 3pi/2
which means that t = pi/6 or t = pi/2
sin t = 0 when t = 0 or pi

26(sin2t)(cos2t)=0 means that
sin2t = 0 or cos2t = 0
sinx = 0 when x = 0 or x = pi, so
sin2t = 0 when 2t = 0 or 2t = pi
So t = 0 or t = pi/2
cosx = 0 when x= pi/2 or x = 3pi/2, so
cos2t = 0 when 2t = pi/2 or 2t = 3pi/2
So t = pi/4 or t = 3pi/4

Namenot is right. I forgot some of the solutions.

Answer 4

-27 cos^2 3t sin t = 0
=> cos 3t = 0 or sin t = 0
=> 3t = (n + 1/2)π or t = nπ, for any n ∈ Z
=> t ∈ {π/6 + nπ/3, n ∈ Z} ∪ {nπ, n ∈ Z}

26(sin2t)(cos2t)=0
=> 13 sin 4t = 0
=> 4t = nπ, n ∈ Z
=> t = nπ/4, n ∈ Z

Answer 5

use substitution
cos2x = cos^2-sin^2
sin2x =2sinxcosx
26(2sinxcosx)(2cos^2x-1)=0
/26&/2
cosxsinx(2cox^2x-1)=0
2cos^3xsinx-cosxsinx=0
sinx(2cos^3x-cosx)=0
sinx*cosx*(2cos^2x-1)=0
sinx=0
cosx=0
2cos^2x-1=0
cos^2x =1/2
cosx =+or -(sqrt 2)/2
sinx = 0 at 0 and 180 or 0 and pi
cosx = 0 at 90 and 270 or pi/2 and 3pi/2
cosx = (sqrt2)/2 at 45 and 315 or pi/4and 7pi/4
cosx = -(sqrt2)/2 at 135 and 225 or 3pi /4 and 5pi/4

Answer 6

Did you try a calculator?