Question

Please help me solve this.

Answer 1

Sin^2(x) + cos^2(x) = 1
sin^2(x) = 1 - cos^2(x)

Replace sin^2(x) in the equation with 1 - cos^2(x)
2(1 - cos^2(x)) - 5cosx + 1 = 0
2 - 2cos^2(x) - 5cosx + 1 = 0
2cos^2(x) + 5cos x - 3 = 0
Let cosx = y
2y^2 + 5y - 3 = 0
(2y -1)(y + 3) = 0
y = cosx so:
cosx = ½,
cosx = -3 (gives no solutions since cos domain is -1 to 1)
Therefore x = pi/3, (2pi - pi/3) = 5/3pi

Answer 2

Answer: x = Π/3

Step-by-step explanation:

From trigonometry identity,

sin²x + cos²x = 1

sin²x = 1 - cos²x ... (1)

Given the equation,

2sin²x−5cosx+1=0 ... (2)

Substituting equation 1 into 2, we will have;

2(1-cos²x)-5cos(x)+1 = 0

2-2cos²x-5cos(x)+1 = 0

-2cos²x-5cos(x)+3 = 0

Multiplying through by minus will give;

2cos²x+5cos(x)-3 = 0

Let P = cos(x)... (3)

The equation will become;

2P²+5P-3 = 0

Factorizing the equation gotten, we have;

2P²+6P-P-3 = 0

2P(P+3)-1(P+3) = 0

(2P-1)(P+3) = 0

2P-1 = 0; P = 1/2

P+3 = 0; P = -3

Therefore P = 1/2 and -3

Substituting the value of P in equation 3, we have;

cos(x) = 1/2

x = arccos1/2

x = 60° = Π/3

Since the second value of P is negative, it gives no solution so we will neglect it.

Therefore, x = Π/3 for 0<x<2Π