Me solve this. solve the given equation in the interval [0,2 π]. note: the answer must be written as a multiple of π. give exact answers. do not use decimal numbers. the answer must be an integer or a fraction. note that π is already provided with the answer so you just have to find the appropriate multiple. e. g. if the answer is π2 you should enter 1/2. if there is more than one answer write them separated by commas. 2(sinx)^2−5cosx+1=0

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

answer: idk

step-by-step explanation:

my guess would be b and d, but i didn’t actually get out the pen and paper. lol