Question

Solve the following equations in the interval [0, 2π]. (Round your answers to 3 decimal places

tan(x)sin(2x) = 1

Answer 1

\[\tan(x)=\frac{sin(x)}{cos(x)}\]
\[sin(2x)=2sin(x)cos(x)\]
use these

Answer 2

sin x * 2 sinx cos x = 1
----
cos x

2sin^2 (x) cos x = cos x
cos x (2sin^2 x - 1) = 0
cos x = 0 or 2sin^2 x = 1

can you proceed from here?

Answer 3

i cant im sry i dont really get this

Answer 4

if cos x = 0 you can find x using your calculator
by entering 0 and pressing either cos-1 key or it might have arcsin key

x will be pi/2 , 3pi/2 radians

sin x^2 = 1/2
sin x = sqrt(1/2) or - sqrt (1/2)

x will be pi/4, 3pi/4, 5pi/4 and 7pi/4 radians

total solution is pi/2 , 3pi/2 , pi/4, 3pi/4, 5pi/4 and 7pi/4 radians

Answer 5

thank you very much

Answer 6

thats ok

hope you've learned something