Find all solutions of the equation

Question

Find all solutions of the equation in the interval [0, 2pi)

4 cos^2 x= 5+4 sin x

write your answer in radians in terms of pi

Answer 1

can solve by ,,,,

4-4sin²x=5+4sinx

4sin²x+4sinx+1=0

sinx=(-1/2)+/-0

x=7π/6,,or x=11π/6

plugging in both cases have 3=3

Answer 2

answer: 7pi/6 and 11pi/6 radians

4cos^2(x) = 5+4sinx

=5 +/-4sqr(1-cos^2(x))

let y = cos^2(x)

4y =5 +/-4sqr(1-y)

isolate the square root term

4y-5 = +/-4sqr(1-y)

4y -5 = +/-sqr(16-16y)

square both sides

16y^2 -40y +25 =16-16y

16y^2 -24y + 9=0

factor or use the quadratic formula if you can't find the factors

(4y-3)^2 = 0

take the square root of both sides

4y-3 = 0

4y = 3

y = 3/4

y=cos^2(x) = 3/4

cosx = +/-sqr3/2

x = 30, 150, 210,330

= pi/6, 5pi/6, 7pi/6 or 11pi/6 radians

but only 7pi/6 and 11pi/6 work

check the answers

4cos^2(330) = 5+4sin330

4(sqr3)/2)^2 = 5+4(-1/2)

4(3/4) = 5-2

3 = 3

30, 150 don't work

330 & 210 work

4cos^2(x) = 5+4sinx

4cos^2(30) = 5+4sin30

4(sqr3/2)^2 = 5+4(1/2)

4(3/4) =3 doesn't = 5.5

try 150 degrees

the left side is negative while the right is positive

try 210=7pi/6

4cos^2(210) = 5+4sin210

4(-sqr3/2)^2 = 5+4(-1/2)

4(3/4) = 5-2

3 =3

this isn't just to check the answer to see if you did the arithmetic or algebra right,

but whenever you multiply by a variable, as when you square both sides, you potentially

introduce extraneous solutions that aren't solutions to the original problem