Solve cos θ+sin θ=cos 2θ+sin 2θ?

2 Answers
Jun 20, 2018

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.

Jun 21, 2018

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