2
![2 \sqrt{3} \sin { \alpha }^{2} - \cos \alpha = 0 2 \sqrt{3} \sin { \alpha }^{2} - \cos \alpha = 0](https://tex.z-dn.net/?f=2+%5Csqrt%7B3%7D+++%5Csin+%7B+%5Calpha+%7D%5E%7B2%7D++-++%5Ccos+%5Calpha+++%3D+0)
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Answer:
2cos
2
x+3sinx=0
⇒2(1−sin
2
x)+3sinx=0
⇒2(sin
2
x)−3sinx−2=0
⇒2(sin
2
x)−4sinx+sinx−2=0
⇒2sinx(sinx−2)+(sinx−2)=0
⇒(sinx−2)+(2sinx+1)=0
⇒sinx=2,sinx=−
2
1
(Not possible)
sinx=−
2
1
=−sin
6
π
=sin(π+
6
π
)
=sin
6
7π
⇒sinx=sin
6
7π
⇒x={nπ+(−1)
n
6
7π
}where n∈I
Hence, the general solution is:
x={nπ+(−1)
n
6
7π
}, n∈I
Step-by-step explanation:
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