How do you solve the equation on the interval #[0,2pi)# for #8 sin^2 theta + 2 sin theta - 1 = 0#?

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Answer 1 · 2017-01-28T07:31:36

The solutions are #S={14.5º , 165.5 º , 210º, 330º}#

You solve this equation like the quadratic equation

#ax^2+bx+c=0#

#8sin^2theta+2sintheta-1=0#

We start by calculating the discriminant

#Delta=b^2-4ac=4-4*8*(-1)=36#

As, #Delta>0#, we have 2 real roots

#x=(-b+-sqrtDelta)/(2a)#

Therefore,

#sin theta=(-2+-sqrt36)/(16)=(-2+-6)/16#

So,

#sintheta=-8/16=-1/2#

#theta=210ª , 330º#

#sintheta=4/16=1/4#

#theta=14.5º , 165.5º#