general solution of sinx=sin9x
Answers
Answered by
4
Square both sides:
(sinx + cosx)^2 = 1^2
sin^2 x + cos^2 x + 2 sinx cosx = 1
1 + 2 sinx cosx = 1
2 sinx cosx = 0
sinx = 0 or cosx = 0
x = 0, pi/2, pi, 3pi/2
Now we must check these values, since we squared both sides at beginning.
x = 0
sin(0) + cos(0) = 0 + 1 = 1
x = pi/2
sin(pi/2) + cos(pi/2) = 1 + 0 = 1
x = pi
sin(pi) + cos(pi) = 0 - 1 = -1
x = 3pi/2
sin(3pi/2) + cos(3pi/2) = -1 + 0 = -1
Only valid answers are x = 0, pi/2
General solution:
x = 2pi*k, pi/2 + 2pi*k, for all integers k
(sinx + cosx)^2 = 1^2
sin^2 x + cos^2 x + 2 sinx cosx = 1
1 + 2 sinx cosx = 1
2 sinx cosx = 0
sinx = 0 or cosx = 0
x = 0, pi/2, pi, 3pi/2
Now we must check these values, since we squared both sides at beginning.
x = 0
sin(0) + cos(0) = 0 + 1 = 1
x = pi/2
sin(pi/2) + cos(pi/2) = 1 + 0 = 1
x = pi
sin(pi) + cos(pi) = 0 - 1 = -1
x = 3pi/2
sin(3pi/2) + cos(3pi/2) = -1 + 0 = -1
Only valid answers are x = 0, pi/2
General solution:
x = 2pi*k, pi/2 + 2pi*k, for all integers k
Sushobit:
have u even read the question i askef
yeah
thats the answer
its the wrong approach
btw ans is wrong
Answered by
1
Answer:
Step-by-step explanation:
Similar questions