How do you solve 4sin^2x+1=-4sinx4sin2x+1=−4sinx?

1 Answer
maganbhai P.
Mar 10, 2018

x=kpi-(-1)^k(pi/6),kinZx=kπ−(−1)k(π6),k∈Z

Explanation:

4sin^2x+4sinx+1=0=>(2sinx+1)^2=04sin2x+4sinx+1=0⇒(2sinx+1)2=0=>2sinx+1=0=>sinx=-1/2=sin(-pi/6)⇒2sinx+1=0⇒sinx=−12=sin(−π6)
=>x=kpi+(-1)^k(-pi/6),kinZ⇒x=kπ+(−1)k(−π6),k∈Z
=>x=kpi-(-1)^k(pi/6),kinZ⇒x=kπ−(−1)k(π6),k∈Z

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