how do you solve trig questions?

Question

how do you solve 3sin^2theta+5cos theta -1= 0

please show wrking out
is there a rule for solving this equation please let me know.fnx

Answer 1

The first answer is very well written out. Basically the rule is that you need to replace terms that don't match the other terms with terms that do, using known identities like sin^x = 1 - cos^2 x so that you can isolate the matching term.

Sometimes the terms don't need to match, as in

sin(2x) + sinx = 0

Here you'd replace the sin(2x) with the known equivalent 2 sinxcosx giving you

2sinx cosx + sinx = 0

then factor out the sinx;
sinx(2cosx + 1) = 0

so sinx = 0 or cosx = -1/2

You just have to look for things that have identities

Answer 2

The first thing I'm going to do is write "x" instead of theta. It just looks cleaner.

3sin^2(x) + 5cos(x) - 1 = 0

Your first step would be to change everything to cosines. Note that there's an identity, sin^2(y) = 1 - cos^2(y). Applying that, we get:

3[1 - cos^2(x)] + 5cos(x) - 1 = 0

Expanding this,

3 - 3cos^2(x) + 5cos(x) - 1 = 0

Grouping like terms,

-3cos^2(x) + 5cos(x) + 2 = 0

And now, multiplying the whole equation by (-1), we get

3cos^2(x) - 5cos(x) - 2 = 0

Note that this is actually a quadratic in disguise. If you don't believe me, let y = cos(x). Then we have

3y^2 - 5y - 2 = 0

At this point, we factor normally, and through trial and error, we should get
(3y + 1) (y - 2) = 0

Therefore, 3y + 1 = 0, y - 2 = 0.
Let's plug y = cos(x) back in, so we get

3cosx + 1 = 0, cosx - 2 = 0

Solving each of these, we get
cos(x) = -1/3, cos(x) = 2

We can reject cos(x) = 2, since -1 <= cos(x) <= 1, therefore we
just have

cos(x) = -1/3

We don't have a unit circle value for this, but provided the solution of x is in the interval [0, 2pi), your two solutions should be

x = {cos^(-1)(-1/3), pi - cos^(-1)(-1/3)}

When I say "cos^(-1)", i mean "cos inverse", i.e. the button on your calculator when pushing the 2nd function button on cos.

Answer 3

let Theta = X
3sin²X + 5cosX - 1 = 0
3(1-cos²X) + 5cosX -1= 0
3 - 3cos²X + 5cosX -1 = 0
3cos²X - 5cosX - 2 = 0
(3cosX + 1)(cosX - 2) = 0
cosX= -1/3 , cosX=2
cosX=-1/3 is the only solution as cosX cannot =2
X must lie in 2nd and 3rd quadrants
use inverse cos on calculator to obtain 71°
X=(180 - 71)° and (180 + 71)°
X=109° and 251°

Answer 4

3sin^2theta+5cos theta -1= 0
use pythagoras
3(1-5cos^(2) theta)+5cos theta
-1=0
3cos^2theta-5costheta-2=0
(costheta-2)(3costheta+1)=0
>>>>costheta=2 or -1/3
costheta lies between
+ and -1,hence 2 is void

therefore, theta=cos^(-1)(-1/3)

since theta is in the 2nd and 3rd
quadrants,
theta=180+or-cos^(-1)(1/3)
=109.4712206 degrees
or 250.5287794 degrees

i hope that this helps

Answer 5

3 sin² Θ + 5 cos Θ - 1 = 0

Here's what you need to know:

sin² Θ = 1 - cos² Θ

Substitute
3 ( 1 - cos² Θ) + 5 cos Θ -1 = 0
3 - 3 cos² Θ + 5 cos Θ - 1 = 0
[abracadabra]
3 cos² Θ - 5 cos Θ -2 = 0
(3 cos Θ + 1) (cos Θ -2) = 0
cosΘ = -1/3 or cosΘ=2.

cos Θ can never =2 so that solution can be excluded
cosΘ = -1/3 is the good root
Θ = arccos (-1/3)

I hope that's right.