# 4cos^2t-1=sin^2t+sint*costFind t !

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may also use the alternative method, hence, you should come up with the substitution, such that:

`tan (t/2) = u`

`sin t = (2u)/(1 + u^2)`

`cos t = (1 - u^2)/(1 + u^2)`

You need to change the variable such that:

`4(1 - u^2)^2/(1 + u^2)^2 - 1 = (4u^2)/(1 + u^2)^2 + (2u(1 - u^2))/(1 + u^2)^2`

Bringing all terms to a common denominator, yields:

`4(1 - u^2)^2 -(1 + u^2)^2 = 4u^2 + 2u - 2u^3`

`(2 - 2u^2 - 1 - u^2)(2 - 2u^2 + 1 + u^2) = 4u^2 + 2u - 2u^3`

`(1 - 3u^2)(3 - u^2) = 4u^2 + 2u - 2u^3`

`3 - u^2 - 9u^2 + 3u^4 = 4u^2 + 2u - 2u^3`

`3u^4 + 2u^3 - 14u^2 - 2u + 3 = 0`

`u_1 = 2/5 => tan(t/2) = 2/5 => t/2 = tan^(-1)(2/5) + n*pi`

`t = 2tan^(-1)(2/5) + 2n*pi`

`u_2 = 9/5 => t = 2tan^(-1)(9/5) + 2n*pi`

`u_3 = -12/5 => t = -2tan^(-1)(12/5) + 2n*pi`

`u_4 = -1/2 => t = -2tan^(-1)(1/2) + 2n*pi`

Hence, evaluating the solutions to the given equation yields `t = 2tan^(-1)(2/5) + 2n*pi ; t = 2tan^(-1)(9/5) + 2n*pi ; t = -2tan^(-1)(12/5) + 2n*pi ; t = -2tan^(-1)(1/2) + 2n*pi.`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To solve for t means to find the angle t from the given identity. We'll transform the given identity into a homogenous equation by substituting 1 by (sin t)^2 + (cos t)^2 = 1 and moving all terms to one side.

(sin t)^2 + sint*cost -  4(cos t)^2 + (sin t)^2 + (cos t)^2 = 0

We'll combine like terms:

2(sin t)^2 + sint*cost - 3(cos t)^2 = 0

Since cos t is different from zero, we'll divide the entire equation by (cos t)^2:

2(sin t)^2/(cos t)^2 + sint*cost/(cos t)^2 - 3 = 0

According to the rule, the ratio sin t/cos t = tan t.

2 (tan t)^2 + tan t - 3 = 0

We'll substitute tan t = x:

2x^2 + x - 3 = 0

x1 = [-1+sqrt(1+24)]/4

x1 = (-1+5)/4

x1 = 1

x2 = (-1-5)/4

x2 = -3/2

We'll put tan t = x1:

tan t = 1

t = arctan 1 + k*pi

t = pi/4 + k*pi

tan t = x2

tan t = -3/2

t = - arctan (3/2) + k*pi

The solutions of the equation are the values of t angle:

{pi/4 + k*pi} U {- arctan (3/2) + k*pi}