Question

2
$2 \sqrt{3} \sin { \alpha }^{2} - \cos \alpha = 0$
​

Answer 1

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: