Question

Arrange the steps in the correct order to solve this trigonometric equation 2sin^2 x -sinx -1 =0 for 0 degrees is less than or equal to x is less than or equal to 90 degrees

Answer 1

Answer:

90°

Step-by-step explanation:

Given the trigonometry equation;

$2sin^2 x -sinx -1 =0$

Step 1; Let p = sin(x) to have;

$2sin^2 x -sinx -1 =0\\2P^2 -P -1 =0$

Step 2: factorize the resulting quadratic equation:

$2P^2 -P -1 =0\\2P^2 -2P+P -1 =0\\\2P(P-1)+1(P-1)=0\\(2P+1)(P-1) = 0\\2P + 1 = 0 \ and \ P-1 = 0\\P = -1/2 \ and \ P =1$

Step 3: Find x:

when p = 1

sin(x) = 1

$x = sin^{-1}1\\x = 90^0$

also when p  -1/2

sin (x) = -1/2

$x = sin^{-1}-1/2\\x = -30^0$

$x = 180-30\\x = 150^0$

Since 150° is not within the range 0<x<90, then the only solution is 90°