Question

Solve cos θ+sin θ=cos 2θ+sin 2θ?

Answer 1

`theta in {2kpi}uu{2/3kpi+pi/6}, k in ZZ`.

Explanation:

`costheta+sintheta=cos2theta+sin2theta`.

`:. ul(cos2theta-costheta)+ul(sin2theta-sintheta)=0`.

`:. -2sin(3/2theta)sin(1/2theta)+2cos(3/2theta)sin(1/2theta)=0`.

`:. 2sin(1/2theta){cos(3/2theta)-sin(3/2theta)}=0`.

`:. sin(1/2theta)=0, or, sin(3/2theta)=cos(3/2theta)`.

`"If, "sin(1/2theta)=0," then, "1/2theta=kpi, k in ZZ,`

`or, theta=2kpi, k in ZZ`.

`"In case, "sin(3/2theta)=cos(3/2theta)............(ast),`

`" then, "cos(3/2theta)!=0, because if "cos(3/2theta)=0,`

`" then "(ast) rArr sin(3/2theta)=0," and this contradicts "`

`sin^2(3/2theta)+cos^2(3/2theta)=0`.

`"So, dividing "(ast)" by "cos(3/2theta)," we get," tan(3/2theta)=1`.

`:. 3/2theta=kpi+pi/4, i.e., theta=2/3kpi+pi/6, k in ZZ`.

Altogether, `theta in {2kpi}uu{2/3kpi+pi/6}, k in ZZ`.

Answer 2

`t = pi/6 + (4kpi)/3`
`t = (5pi)/6 + (4kpi)/3`
`t = 4kpi`
`t = (k + 1)2pi`

Explanation:

sin 2t + cos 2t = sin t + cos t
Use trig identity; `sin a + cos a = sqrt2sin (a + pi/4)`.
In this case:
`sqrt2sin (2t + pi/4) - sqrt2sin(t + pi/4) = 0`
`sin (2t + pi/4) - sin (t + pi/4) = 0` (1)
Use trig identity:
`sin a - sin b = 2cos ((a + b)/2)sin ((a -b)/2)`
`(a + b)/2 = (3t)/2 + pi/4`
`(a - b)/2 = t/2`
Equation (1) becomes:
`-2cos ((3t)/2 + pi/4).sin (t/2) = 0`
Either factor should be zero.
a. `cos ((3t)/2 + pi/4) = 0`
1. `(3t)/2 + pi/4 = pi/2` --> `(3t)/2 = pi/2 - pi/4 = pi/4 + 2kpi`
`3t = pi/2 + 4kpi` --> `t = pi/6 + (4kpi)/3`
2. `(3t)/2 + pi/4 = (3pi)/2` --> `(3t)/2 = (5pi)/4 + 2kpi`
`3t = (5pi)/2 + 4kpi` --> `t = (5pi)/6 + (4kpi)/3`
b. `sin (t/2) = 0`
1. `t/2 = 2kpi` --> `t = 4kpi`
2. `t/2 = pi` --> `t = 2pi + 2kpi = (k + 1)2pi`