How can i solve this equation: 2sen^2(t)-cos(t)-1=0 ?

3 Answers
Sonnhard
May 20, 2018

If you meant #sin^2(t)# then you can use #sin^2(t)=1-cos^2(t)# and you will get (after simplifications) #cos^2(t)+1/2cos(t)-1/2=0#

Explanation:

After applying the hint you will get #2(1-cos^2(t))-cos(t)-1=0#
This is
#cos^2(t)+1/2cos(t)-1/2=0#
substituting #z=cos(t)# you have to solve #z^2+1/2z-1/2=0#

Nghi N
May 20, 2018

#t = (2k+1)pi#
#t = +- p/3 + 2kpi#

Explanation:

#2sin^2 t - cos t - 1 = 0#
Replace #sin^2 t# by #(1 - cos^2 t)#
#2 - 2cos^2 t - cos t - 1 = 0#
Solve this quadratic equation for cos t:
#-2cos^2 t - cos t + 1 = 0#
Since a - b + c = 0, use shortcut. The 2 real roots are:
#cos t = - 1#, and #cos t = -c/a = 1/2#
a. cos t = -1
Unit circle gives --># t = pi + 2kpi = (2k+ 1)pi#
b. #cos t = 1/2#
Trig table and unit circle give -->
#t = +- pi/3 + 2kpi#

A. S. Adikesavan
Aug 10, 2018

An odd multiple of #pi and ( 6k +- 1)pi/3, k = 0, +- 1, +-2, +-3, ...#

Explanation:

using # sin^2t = 1 - cos^2t#, te equation becomes a quadratic in

#cost#, giving

cos t = -1, 1/2 = cos pi, cos pi/3# and these give

#t = 2kpi +- pi and 2kpi +- pi/3, k = 0, 1, 2, 3, ...#

#= an odd multiple of pi, ( 6k +- 1)pi/3.

The list near 0 is

#t = .....- 7/3pi, - 5/3pi,- pi, - pi/3, pi/3, pi, 5/3pi, ...

See graph, for t-intercepts as solutions.
graph{y-2 (sin(x))^2 +cos (x) + 1=0[-10 10 -5 5]}

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