solve the equation: 4cos2*theta + 3cos*theta + 3 = 0 for values between 0 and 180 degrees?

Question

i typed the first question in wrong. Above is how it should read.

Answer 1

4cos2x +3cosx +3 = 0 [x=theta]
Let y=2x
Then 4cos y +3 cos(y/2) +3 = 0
Now use the cos 1/2 angle formula to get:
4cos y +3 [sqrt (1+cosy)/2] +3 =0
3 [sqrt (1+cos y)/2] = -3 -4cos y
9(1+cos y)/2= 9 +24 cos y +16cos^2y
9+9cosy =18 +48cos y +32cos^2y
32cos^2y +39cos y +9 =0
cos y = [-39 + or - sqrt(39^2- 4*32*9)]/2
cos y = -39/2 + 3/2sqrt (41), and
cos y = -39/2 -3/2sqrt(41)
y=arccos[-39/2 + 3/2sqrt (41)] , and
y=arccos[-39/2 - 3/2sqrt (41)]
x= y/2 =1/2*arccos[-39/2 + 3/2sqrt (41)], and
x=1/2*arccos[-39/2 - 3/2sqrt (41)]
Use your calculator to find final answer

Answer 2

3 sin2x = cos2x => tan2x = a million/3 ; x =9d 13' 3" {If i'm incorrect sin2x=> sq. of sinx, then x = 60d} a million. 3/5 = tanx => x = 30d fifty seven' 50" 2. (a)sin 30=a million/2; cos 30=sqrt3/2 (b)tan 30= sin 30/cos 30 = a million/sqrt3 3. enable a = BC, b = AC=sqrt3/2, c = AB=6 b^2 = c^2 + a^2 - 2ac cos(B) 3/4 = 36 + a^2 -12 a(sqrt3/2) a^2 -6 sqrt3 a + 35.25 =0 a = 3 sqrt3 [+/-]sqrt[27-35.25] because the novel supplies an imaginary quantity the answer isn't plausible 4. (a)2 cos^2 x + sin x =2[a million-sin^2 x] + sin x =2 + sin x - 2sin^2 x (b)If the above expression is =2 =>sin x - 2 sin^2 x =0 => sin x(a million -2 sin x)= 0 =>(a million/2 -sin x) =0 => sin x =a million/2 & x=30. 5. (a)3 sin^2 x + 4 cos x =3[a million - cos^2 x] + 4 cos x = -3 cos^2 x + 4 cos x + a million (b) 3[a million - cos^2 x] + 4 cos x - 4 =0 -3 cos^2 x + 4 cos x - a million =0; cos x = 2/3 [+/-]sqrt[4/9 -a million/3] =2/3[+/-]sqrt[a million/9] = a million, a million/3 x = 90d, 70d 31'40 3.sixty one"

Answer 3

4cos2(theta)+3cos(theta)+3=0
take theta =x
so 4cos2x+3cosx+3=0
put cos2x=2cos^x-1
4(2cos^x-1)+3cosx+3=0
8cos^x-4+3cosx+3=0
8cos^x+3cosx-1=0
this is a quadratic equation solve for x and u will get value of theta

Answer 4

The answer is 77.77 degree

Allahuakbar