Question

Question #dcedf

Answer 1

`theta_1 ~~ .333 and theta_2~~1.238`
Solution obtained Graphically... see below and explanation

Solution set on `(0,pi) => (.333, 0) and (1.238,0)`

Explanation:

Given: `co2theta = tan2theta " "` `theta: theta in 0"<"theta"<"pi`

Required: The solution to `=>r(theta)=co2theta - tan2theta`
over `theta: theta in(0, pi )`

Solution Strategy:
a) Use trigonometric identity - `tan2theta= (sin2theta)/(cos2theta)`
b) Solve the rational function resulting from a) by any means necessary...

a) Substituting `tantheta= sintheta/costheta` we write:
`cos2theta = (sin2theta)/(cos2theta)`
`cos^(2)2theta = sin2theta` now this leads to
1) `r(theta)=cos^(2)2theta -sin2theta=0` find the roots
substitute `sin2theta = sqrt[1-cos^(2)2theta]`
2) `r(theta)= cos^(2)2theta - sqrt[1-cos^(2)2theta]`

b) Solve 1) form directly or 2) form indirectly
I have chosen to so solve form 1) graphically see below:
Over `theta:theta (0,pi)` we have 2 solutions
`theta_1 ~~ .333 and theta_2~~1.238`

You also cam graph, #r(theta)=cos^(2)2theta -sin2theta and locate the zeros, x intercept...

Get the same answer (.333, 0) and (1.238,0)

Good luck!

Answer 2

`19^@09 and 70^@92`

Explanation:

`cos 2t = (sin 2t)/(cos 2t)`
Cross multiply -->
`cos^2 2t - sin 2t = 0`
`(1 - sin^2 2t) - sin 2t = 0`. --> Trig identity: `cos^2a + sin^2 a = 1`
`- sin^2 2t - sin 2t + 1 = 0`
Solve this quadratic equation for sin 2t.
`D = d^2 = b^2 - 4ac = 1 + 4 = 5` --> `d = +- sqrt5`
There are 2 real roots:
`sin 2t = -b/(2a) +- d/(2a) = - 1/2 +- sqrt5/2 = -(1/2)((1 +- sqrt5)`
a. `sin 2t = -(1 + sqrt5)/2 = - 3.23/2 = - 1.62` (Rejected since < -1)
b. `sin 2t = - (1 - sqrt5)/2 = 0.62`
sin 2t = 0.62. Calculator gives -->`2t = 38.17^@ --> t = 19^@09`
Trig unit circle gives another arc 2t that has the same sin value
`2t = 180 - 38.17 = 141^@83 --> x= 70^@92`
Answer for `(0, pi):`
1`9^@09; 70^@92`