How do you solve sin2θ=-cosθ ?

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Answer 1 · 2018-04-27T17:31:59

#theta={(2k+1)pi/2,kinZZ}uu{kpi-(-1)^k*pi/6,kinZZ}#

We know that,

#color(red)(sinx=sinalpha=>x=kpi+(-1)^k*alpha,kinZZ#

Here,

#sin2theta=-costheta#

#=>sin2theta+costheta=0#

#=>2sinthetacostheta+costheta=0#

#=>costheta(2sintheta+1)=0#

#=>costheta=0 or 2sintheta+1=0#

#=costheta=0 or sintheta=-1/2#

#(i)color(blue)(costheta=0=>theta=(2k+1)pi/2,kinZZ#

#(ii)sintheta=-1/2=>sintheta=sin(-pi/6)#

#=>theta=kpi+(-1)^k(-pi/6),kinZZ#

#=>color(blue)(theta=kpi-(-1)^k(pi/6),kinZZ#

Hence,

#theta={(2k+1)pi/2,kinZZ}uu{kpi-(-1)^k*pi/6,kinZZ}#

Answer 2 · 2018-04-27T17:32:10

Within the interval of #0 < theta < 2pi#

#theta=pi/2, (3pi)/2, (7pi)/6, (11pi)/6#

.

We use the double-angle formula:

#sin2theta=-costheta#

#2sinthetacostheta=-costheta#

We move all terms to one side.

#2sinthetacostheta+costheta=0#

We factor out #costheta#

#costheta(2sintheta+1)=0#

Then, we set each piece equal to #0# and solve for #theta#.

#costheta=0, :. theta=pi/2, (3pi)/2#

#2sintheta+1=0#

#sintheta=-1/2, :. theta=(7pi)/6, (11pi)/6#