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3sin^2theta+7sinsintheta +2=0
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sinsin?
I answered this already?
would u anser again
replace sin(x) with a varaible, then solve the resulting quadratic
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*sin(theta)
after replacing sin theta with a, we get: 3a^2+7a+2=0
after that
well you gotta factor it, you should get: (a+2)(3a+1)=0
Then solve for a ineach term: a+2=0----->a=-2 3a+1=0---->a=-1/3
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then replace a with sin(theta)
but\[-1 \le \sin \theta \le 1\] so neglect the a=-2 and solve for theta in sin(theta)= -1/3
Nice job @lalaly
lol thankyou :$
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