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grin007 [14]
3 years ago
8

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 e

qual to x is less than or equal to 90 degrees
Mathematics
1 answer:
liq [111]3 years ago
7 0

Answer:

<h3>90°</h3>

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°

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