Solve the equation #cos2theta+sintheta=0#, if #0<=theta<=2pi#?

2 Answers
Harold Walden
Jan 7, 2018

#theta=pi/2,(7pi)/6, (11pi)/6#

Explanation:

#cos2θ + sinθ = 0#
#cos(2theta)=1-2sin^2theta#
#:. cos2θ + sinθ=1-2sin^2theta+sintheta=0#
(rearranging to look more like a quadratic)
#-2sin^2theta+sintheta+1=0#
#color(red)(-1)(-2sin^2theta+sintheta+1)=color(red)(-1)xx0#
#2sin^2theta-sintheta-1=0#
Let #x=sintheta#
#2x^2-x-1=0#
Factorising the quadratic yields;
#(2x+1)(x-1)=0#
By the null factor law;
#x=-1/2# or #x=1#
#:. sintheta=-1/2# or #sintheta=1#
Hence,
#theta=pi/2,(7pi)/6, (11pi)/6# on the domain #[0,2pi]#

I hope that helps :)

Nam D. · Shwetank Mauria
Jan 7, 2018

#theta=pi/2# or #(7pi)/6# or #(11pi)/6#

Explanation:

We have

#cos2theta+sintheta=0#, solutions in #0<=theta<=2pi#

Substitute #cos^2theta=1-2sin^2theta#

#1-2sin^2theta+sintheta=0#

Let #sintheta=u#

#1-2u^2+u=0#

#<=>2u^2-u-1=0#

or #(2u+1)(u-1)=0#

#u=1, u=-1/2#

Substitute back #sintheta=1# or #-1/2#

and for #0<=theta<=2pi#

#theta=pi/2# or #pi+pi/6=(7pi)/6# or #2pi-pi/6=(11pi)/6#

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