Answer:
sin(2x)=cos(π2−2x)
So:
cos(π2−2x)=cos(3x)
Now we know that cos(x)=cos(±x) because cosine is an even function. So we see that
(π2−2x)=±3x
i)
π2=5x
x=π10
ii)
π2=−x
x=−π2
Similarly, sin(2x)=sin(2x−2π)=cos(π2−2x−2π)
So we see that
(π2−2x−2π)=±3x
iii)
π2−2π=5x
x=−310π
iv)
π2−2π=−x
x=2π−π2=32π
Finally, we note that the solutions must repeat every 2π because the original functions each repeat every 2π. (The sine function has period π so it has completed exactly two periods over an interval of length 2π. The cosine has period 23π so it has completed exactly three periods over an interval of length 2π. Hence, both functions repeat every 2π2π2π so every solution will repeat every 2π.)
So we get ∀n∈N
i) x=π10+2πn
ii) x=−π2+2πn
iii) x=−310π+2πn
(Note that solution (iv) is redundant since 32π+2πn=−π2+2π(n+1).)
So we conclude that there are really three solutions and then the periodic extensions of those three solutions.
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