Trigonometry Calculus?

Question

Solve for x in [-pi,pi]: tan(x)-sin(2x)=0

thanks

Answer 1

tanx - sin(2x) =

sinx/cosx - 2sinxcosx=

(sinx - 2sinxcosxcosx)/cosx=

(sinx-2sinxcos^2x)/cosx = 0

sinx - 2sinxcos^2x = 0

sinx(1-2cos^2x) = 0

sinx = 0 or 1-2cos^2x = 0

x = 0 + n*pi with n = 0,1,2,..

or

2cos^2x = 1

cos^2x = 1/2

cosx = +- sqrt(2)/2

.x = pi/4 + 2n*pi

Answer 2

♠Let t=tan(x), then sin(2x) = 2t/(1+t^2);
♣Thus t –2t/(1+t^2) = t*(t^2+1 -2) /(1+t^2) =0;
t*(t^2 –1) =0; thus
♥ t1=-1, t3=+1, t5=0; hence
t1 gives x1=-pi/4, x2=-pi/4 +pi =3pi/4;
t3 gives x3=pi/4, x4=pi/4 -pi = -3pi/4;
t5 gives x5=-pi, x6=0, x7= pi;
all in all 7 roots; your teacher should enjoy this!

Answer 3

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Answer 4

sin(x)/cos(x) - 2sin(x)cos(x) = 0
either sin(x) = 0 (x = -pi, 0, pi) or
1/cos(x) -2cos(x) = 0
cos^x = 1/2
or x = arccos(2^-1/2)