Question

If 2sin²A+3sinA=0, find the permissible value of cosA

Answer 1

Answer:

√5/2

Step-by-step explanation:

2sin^2+3sinA=0

2sin^2A=-3sinA

2sinAsinA= -3sinA

2sinA=-3

sinA=-3/2

sin in opposite upon hypotenuse

there fore through Pythagoras theorem

2^2=-3^2+x^2

4=9+x^2

x^2=-5

x=-√5

since

cos is adjacent upon hypotenuse

cosA =-√5/2

Answer 2

EDIT: I think I might have made a mistake, but I can't correct it at the moment.

$2\sin^2\alpha+3\sin \alpha=0\\\sin \alpha(2\sin \alpha+3)=0\\\sin \alpha=0 \vee 2\sin \alpha+3=0\\\alpha=k\pi \vee \sin \alpha=-\dfrac{3}{2}\\\alpha=k\pi \vee \alpha\in \emptyset\\\alpha=k\pi\\\\\sin^2 \alpha +\cos^2 \alpha =1\\\sin^2 k\pi +\cos^2 \alpha =1\\0+\cos^2 \alpha =1\\\cos^2 \alpha=1\\\boxed{\cos \alpha=1 \vee \cos \alpha=-1}$