Answer:
90°
Step-by-step explanation:
Given the trigonometry equation;
[tex]2sin^2 x -sinx -1 =0[/tex]
Step 1; Let p = sin(x) to have;
[tex]2sin^2 x -sinx -1 =0\\2P^2 -P -1 =0[/tex]
Step 2: factorize the resulting quadratic equation:
[tex]2P^2 -P -1 =0\\2P^2 -2P+P -1 =0\\\2P(P-1)+1(P-1)=0\\(2P+1)(P-1) = 0\\2P + 1 = 0 \ and \ P-1 = 0\\P = -1/2 \ and \ P =1[/tex]
Step 3: Find x:
when p = 1
sin(x) = 1
[tex]x = sin^{-1}1\\x = 90^0[/tex]
also when p -1/2
sin (x) = -1/2
[tex]x = sin^{-1}-1/2\\x = -30^0\\[/tex]
[tex]x = 180-30\\x = 150^0[/tex]
Since 150° is not within the range 0<x<90, then the only solution is 90°
