Question

Find the solution of equation 4sin^2 (theta) - 1 = 0 (in the interval [0, pi/2]

Answer 1

bring the 1 on the right hand side you get 4sin(theta)^2 = 1 then divide both side by 4 you get sin(theta)^2 = 1/4 then sqrt the (1/4) and you get sin(theta) = sqrt(1/4) and then take the sin inverse if that to get theta = sin^-1(sqrt(1/4))
Just remember you are trying to figure out what value of theta will make 4sin^2(theta) a 1 because you need 1-1 and then just move everything around like you would with any other algebra problem

Answer 2

oh right I forgot interval umm you also need to convert that answer to radians by (pi/180) and then you get pi/6 and then you add pi to pi over six until you have covered the interval but pi/6 turns out to be the only value because you go over when you add pi I think thats how you do it I may be wrong so let me know if I got it wrong