Solve 5sinθ+2cosθ=5?

3 Answers
P dilip_k
Jun 23, 2018

#5sinθ+2cosθ=5#

#=>2cosθ-5(1-sintheta)=0#

#=>2(cos^2(θ/2)-sin^2(theta/2))-5(cos^2(theta/2)+sin^2(theta/2)-2cos(theta/2)sin(theta/2))=0#

#=>2(cos^2(θ/2)-sin^2(theta/2))-5(cos(theta/2)-sin(theta/2))^2=0#

#=>(cos(theta/2)-sin(theta/2))(2cos(theta/2)+2sin(theta/2)-5cos(theta/2)+5sin(theta/2))=0#

#=>(cos(theta/2)-sin(theta/2))(7sin(theta/2)-3cos(theta/2))=0#

When

#(cos(theta/2)-sin(theta/2))=0#

#=>tan(theta/2)=1=tan(pi/4)#

#=>theta/2=npi+pi/4" where "n in ZZ#

#=>theta=2npi+pi/2" where "n in ZZ#

When

#7sin(theta/2)-3cos(theta/2)=0#

#=>tan(theta/2)=3/7=tanalpha# (say)

#=>theta/2=kpi+alpha" where " k in ZZ#

#=>theta=2kpi+2alpha" where " k in ZZ# and #alpha=tan^-1(3/7)#

Nghi N.
Jun 23, 2018

#t = 46^@63 + k360^@#
#t = 90^@ + k360^@#

Explanation:

5sin t + 2cos t = 5
Divide both sides by 5:
#sin t + (2/5)cos t = 1# (1)
Call #tan a = sin a/(cos a) = 2/5# --> #a = 21^@80#, and cos a = 0.93.
The equation (1) becomes;
#sin t cos a + sin a.cos t = cos a #
#sin (t + 21^@80) = 0.93#
Calculator, unit circle and property of sin function give 2 solutions for (t + 21.80):
#(t + 21.80) = 68.43^@#, and
#(t + 21.80) = 180 - 68.43 = 111^@#57#.

a. t + 21.80 = 68.43 +k360
#t = 46^@63 + k360^@#
b. t + 21.80 = 111.57 +k360
#t = 89^@77 + k360^@#, or --> #t = 90^@ + k360^@#
Check by calculator.
t = 46.63 --> 5sin t = 5(0.726) = 3.63 --> 2cos t = 2(0.686) = 1.37
5sin t + 2cos t = 3.63 + 1.37 = 5. Proved
t = 90^@ --> 5sin t = 5(1) = 5 -> cos t = 0.
5sin t + 2cos t = 5 + 0 = 5. Proved

Ratnaker Mehta
Jun 23, 2018

#x in {kpi+(-1)^kpi/2}uu{kpi+(-1)^karcsin(21/29)}, k in ZZ#.

Explanation:

Given that, #5sintheta+2costheta=5#.

#:. 2costheta=5(1-sintheta)#.

#"Squaring, "4cos^2theta=25(1-sintheta)^2#,

# i.e., 4(1-sin^2theta)=25(1-sintheta)^2#

#"Transposing, "4(1-sintheta)(1+sintheta)-25(1-sintheta)^2=0#.

#:. (1-sintheta){4(1+sintheta)-25(1-sintheta)}=0#,

# or, (1-sintheta)(29sintheta-21)=0#.

#:. sintheta=1, or, sintheta=21/29.#

#sintheta=1=sin(pi/2) rArr theta=kpi+(-1)^kpi/2, k in ZZ#.

#sintheta=21/29=sin(arcsin(21/29))," gives, "#

#theta=kpi+(-1)^karcsin(21/29), k in ZZ#.

Altogether, we get ,

#x in {kpi+(-1)^kpi/2}uu{kpi+(-1)^karcsin(21/29)}, k in ZZ#.

Related questions
Impact of this question
10322 views around the world
You can reuse this answer
Creative Commons License