Find the value of x from sin⁡x+cos⁡x=2sin⁡(5x)I used auxiliary argument method and converted into cos⁡(π4−x)=cos⁡(π2−5x) (by introducing 2 as auxiliary argument and then using cos⁡(π4)cos⁡x+sin⁡(π4)sin⁡x=cos⁡(π4−x) and using sin⁡5x=cos⁡(π2−5x)But the answer isn't matching after using the formula for cos⁡θ=cos⁡α

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Find the value of x from sin⁡x+cos⁡x=2sin⁡(5x)I used auxiliary argument method and converted into cos⁡(π4−x)=cos⁡(π2−5x) (by introducing 2 as auxiliary argument and then using cos⁡(π4)cos⁡x+sin⁡(π4)sin⁡x=cos⁡(π4−x) and using sin⁡5x=cos⁡(π2−5x)But the answer isn&#039;t matching after using the formula for cos⁡θ=cos⁡α

Travis Hicks · Accepted Answer

Then  π4−x=2πk±(π2−5x)  π4−x=2πk+π2−5x   or   π4−x=2πk−π2+5x  4x=2πk+π4   or   −6x=2πk−3π4  x=π(8k+1)16   or   x=−π(8k−3)24   =π(8n+3)24

vicki331g8 · Accepted Answer

It is much simpler to use congruences.   First note the equation can be written as  sin⁡(x+π4)=sin⁡5x⇔{5x≡x+π4 mod 2π⇔4x≡π4 mod 2πor5x≡π−(x+π4) mod 2π⇔6x≡3π4 mod 2π

nick1337 · Accepted Answer

A standard trig formula is sin⁡(a+b)=sin⁡acos⁡b+sin⁡bcos⁡a We can use this to say sin⁡x+cos⁡x=2(sin⁡x×12+cos⁡x×12)=2(sin⁡xcos⁡(π4)+cos⁡xsin⁡(π4))=2sin⁡(x+π4) So 2sin⁡(x+π4)=2sin⁡(5x) Therefore, x+π4=5x (modulo 2π). And you should be able to take it from there. Peterwhy&#039;s comment reminds me, you also need to take into account that

Find the value of x from sin⁡x+cos⁡x=2sin⁡(5x)I used auxiliary argument method and converted into cos⁡(π4−x)=cos⁡(π2−5x)...

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