Trig problem help?

Question

How do I find all the values of theda between 0

Answer 1

First use trig identities so each term has the same form

cos^2(t) = cos^2(t) - sin^2(t) = 1 - 2sin^2(t)

so 3cos(2t) + sin(t) - 1 = 0 becomes

3 - 6sin^2(t) + sin(t) - 1 = 0

Let x=sin(t)

6x^2 - x -2 = 0

Now solve the quadratic equation

a=6, b=-1, c=-2

x=[1+/-sqrt(1+48)]/12

x=(1+/-7)/12 --> x=2/3 x=-1/2

so the solutions are sin(t)=2/3 and sin(t)=-1/2

So what values of t (or theda) would give those values?

sin(41.81) = 2/3 sin(138.19) = 2/3
sin(330) = -1/2 sin(210) = -1/2

Answer 2

First I will set theta = x because it is easier.

3 cos 2x + sin x -1 = 0

cos 2x = cos^2 x - sin^2 x, so I will plug this into the equation

3 {cos^2 x - sin^2 x} + sin x - 1 = 0

I will now use the trig identity

cos^2 x + sin^2 x = 1

3{1 - sin^2 x - sin^2 x} + sin x - 1 = 0

3 - 6 sin^2 x + sin x - 1 = 0

-6 sin^2 x + sin x + 2 = 0

I will make the substitution z = sin x

-6 z^2 + z + 2 = 0

Now I will solve for z using either factoring or the quadratic formula. I factor this to:

( 3 z - 2) (-2 z - 1) = 0

which means,

3 z - 2 = 0 or -2 z - 1 = 0

z = 2/3 or -1/2, since z = sin x

sin x = 2/3 or -1/2

sin^-1 (2/3) = x = 41.8 and 138.2 degrees

sin^-1 (-1/2) = x = 210 and 330 degrees

I would check all my math to ensure that I am correct.

Answer 3

For ease of typing I'm going to use x instead of theta.
3cos(2x) + sin(x) - 1 = 0
Using the double angle cosine identity:
3(1-2sin^2(x)) + sin(x) - 1 = 0
3 - 6sin^2(x) + sin(x) - 1 = 0
-6sin^2(x) + sin(x) + 2 = 0 Now let's let sin(x) = u.
-6u^2 + u + 2 = 0 Now it's just a quadratic equation
let's factor:
(-3u + 2)(2u + 1) = 0
u = 2/3 and u = -1/2
So sin(x) = 2/3 and sin(x) = -1/2
to solve we'll use the inverse sine.
arcsin(2/3) = 41.810.
Now remember in the unit circle there will be two places where sin(x) = -1/2 so to find the other place we have to subtract the answer from 180.
180 - 41.810 = 138.189.
Now for the other answer:
sin(x) = -1/2
arcsin(-1/2) = x
arcsin(-1/2) = -30 or 330.
Same concept here, there are two answers.
180 - (-30) = 210.

Here are the answers:
210
330
41.810 = arcsin(2/3)
138.189 = 180 - arcsin(2/3)

If you decide to check them (which I already have) use the inverse sine function because 41.810 and 138.189 are rounded.

Answer 4

I read your problem as cos² (theta) rather than cos (2theta). Doing it this way, I have three answers: 90°, 221.81°, or 318.19°.

Using x to represent theta,
3 cos² x + sin x - 1 = 0
3 (1 - sin² x) + sin x - 1 = 0
3 - 3sin² x + sin x - 1 = 0
-3sin² x + sin x + 2 = 0. Factoring,
(-sin x + 1)(3sin x + 2) = 0.
-sin x + 1 = 0 or 3sin x + 2 = 0.
On the left, sin x = 1, so x = 90°.
On the right, 3sin x = -2, sin x = -2/3, so
x = arcsin(-2/3). This is close to -41.81°, and in your domain,
x is approximately 318.19° or 221.81°.

Answer 5

let theda = x
3cos2x + sin x - 1 = 0
3( 1 - 2 sin^2 x) + sin x - 1 = 0
3 - 6 sin^2 x + sin x - 1 = 0
- 6sin^2 x + sinx + 2 = 0
divide both sides by -1
6sin^2x - sin x - 2 =0
( 3 sin x - 2)(2 sinx + 1) = 0
Either 3sinx - 2=0
sinx = 2/3
x = sin inverse (2/3)
x= 41.8 , 138.2

or 2sinx + 1 = 0
sinx = -1/2
x = sin inverse (1/2)
x = 210, 330

Answer 6

Well lets see: u may want to convert to a standard quadratic eqn, of the form x^2 + x + c = 0
3cos2α + sinα -1 = 0

from trig laws: 3(cos^2 α- sin^2 α) + sinα-1=0
3([1-sin^2 α]-sin^2α)+sinα-1=0
3-3sin^2 α-3sin^2 α+sinα-1=0 (do the algebra, u get:)
6sin^2 α-sinα-2=0

assume sinα=x...and solve the quadratic:

sinα=2/3 and -1/2: decide from there what ur values are

Answer 7

Let "theta" = x and reading cos(2theta) as cos 2x as opposed to cos²x:-
3.(1 - 2.sin² x) + sin x - 1 = 0
- 6.sin² x + sin x + 2 = 0
6.sin² x - sin x - 2 = 0
(3sin x - 2).(2 sin x + 1) = 0
sin x = 2/3, sin x = - 1/2
x = 41.8° , 138.2° , 210° , 330°

Answer 8

3 cos2t + sint - 1 = 0

cos 2t = 1 - 2 (sin^2 ( t ) )

3 - 6 (sin^2 ( t ) ) + sin t - 1 = 0

sin t = x

- 6 x^2 + x + 2 = 0

6 x^2 - x - 2 = 0

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

x = 2 / 3 or x = - 1 / 2

sin t = 2 / 3 -----> t = arcsin (2 / 3) = 41 or 139 (degree)

sin t = -1 / 2 ------> t = 210 or 330 (degree)

Answer 9

"theda" is too long
3cos(2x) +sin (x) -1= 0
3 (1-2sin^2(x)) +sin (x) -1=0
3- 6sin^2(x) + sin(x) -1=0
2- 6sin^2(x) + sin(x)=0
replace sin(x) with q
-6q^2 +q +2 =0
use quadratic formula or whatever to get
q= (-.5) and (2/3)
insert sin(x) into q
sin (x)= -.5 and (2/3)
x= 41.81
x= 138.19
x= 210
x= 330
Replace x with theta

Answer 10

2/(a million+cosx) - tan²(x/2) = a million Taking LHS 2/(a million+cosx) - (sin²x/2 / cos²x/2) 2/(2cos²x/2-a million +a million) -(sin²x/2 / cos²x/2) a million/cos²x/2 -(sin²x/2/cos²x/2) a million-sin²x/2 / cos²x/2 cos²x/2 / cos²x/2 a million (Proved ) Trignometric Identities Used :- cos2x = 2cos²x -a million sin²x + cos²x = a million Tanx = sinx / cosx