# help: trig question (1 Viewer)

#### +GriM ReApeR+

hey pplz , can u solve this 4 me??

Solve 3cos(2x) - 2sin(2x) = 1 using "t method" and "subsidiary method"

thanks!

#### CM_Tutor

##### Moderator
Moderator
Were you asked to solve this over a particular domain (ie 0 <= x <= 2 * pi, for example), or over all real x, as the answers will be different?

#### +GriM ReApeR+

Originally posted by CM_Tutor
Were you asked to solve this over a particular domain (ie 0 <= x <= 2 * pi, for example), or over all real x, as the answers will be different?
yea... it was 0 <= x <= 2(pie)

#### CM_Tutor

##### Moderator
Moderator
t - method:

Let t = tan x. Then cos 2x = (1 - t^2) / (1 + t^2) and sin 2x = 2t / (1 + t^2).
Substituting, we get 3(1 - t^2) / (1 + t^2) - 2(2t) / (1 + t^2) = 1
which simplifies to 2t^2 + 2t - 1 = 0
which solves to t = (-1 - sqrt(3)) / 2 or (sqrt(3) - 1) / 2

The first of these gives solutions of x = n * pi - alpha, where alpha = 0.9388 ... and n is an integer. In the required domain, this is x = 2.203 (when n = 1) or x = 5.344 (when n = 2) (4 sig fig).

The second of these gives solutions of x = n * pi + beta, where beta = 0.3508 ... and n is an integer. In the required domain, this is x = 0.3509 (when n = 0) or x = 3.492 (when n = 1) (4 sig fig).

Now, t - formulae work provided t is defined, and thus x = pi / 2 and x = 3 * pi / 2 must be checked manually - neither is a solution in this case:

So, x = 0.3509, 2.203, 3.492 or 5.344 (4 sig fig).

#### +GriM ReApeR+

thanks CM_Tutor ^^

#### CM_Tutor

##### Moderator
Moderator
Auxilliary angle method:

Let 3cos 2x - 2sin 2x = Rcos(2x + alpha), where R > 0 and alpha is an acute angle.

Expanding, we get: 3cos 2x - 2sin 2x = R(cos 2x * cos alpha - sin 2x * sin alpha)
Equating coefficients of cos 2x: 3 = Rcos alpha ______ (1)
Equating coefficients of sin 2x: -2 = -Rsin alpha ______ (2)

(1)^2 + (2)^2: 9 + 4 = R^2 * (cos^2 alpha + sin^2 alpha) = R^2
So, R = sqrt(13, as R > 0

(2) / (1): -Rsin alpha / Rcos alpha = -2 / 3
tan alpha = 2 / 3
alpha = 0.58002 ..., as alpha is acute

So, 3cos 2x - 2sin 2x = sqrt(13) * cos(2x + 0.58002...)

Thus, we can transform our original problem, 3cos 2x - 2sin 2x = 1, 0 <= x <= 2 * pi
into sqrt(13) * cos(2x + 0.58802...) = 1, 0 <= 2x <= 4 * pi
and so cos(2x + 0.58802...) = 1 / sqrt(13), 0.58802... <= 2x + 0.58802... <= 13.154...

Let beta be the related angle, acute, satisfying cos beta = 1 / sqrt(13)
Thus, beta = 1.2897...

Since cos(2x + 0.58802...) is positive, we need solutions in the first and fourth quadrants.

So, 2x + 0.58802... = beta, 2 * pi - beta, 2 * pi + beta, 4 * pi - beta in the desirred domain.
So, 2x + 0.58802... = 1.2897..., 4.9934..., 7.5729..., 11.276...
2x = 0.70175..., 4.4054..., 6.9849..., 10.688...
x = 0.35087..., 2.2027..., 3.4924..., 5.3443...

So, x = 0.3509, 2.202, 3.492, 5.344 (4 sig fig).

#### CM_Tutor

##### Moderator
Moderator
Bringing this to the top for Nelly_04.