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 $theta$ .

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

$2sintheta+1=0$

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