Solve for x: 1+\tan^2(x)=8\sin^2(x),\ x\in[\frac{\pi}{6},\frac{\pi}{2}]

abiejose55d

abiejose55d

Answered question

2022-04-21

Solve for x: 1+tan2(x)=8sin2(x), x[π6,π2]

Answer & Explanation

Diya Bass

Diya Bass

Beginner2022-04-22Added 20 answers

Recall 1+tan2(x)=sec2(x) and sin(2x)=2sin(x)cos(x). Hence,
1+tan2(x)=8sin2(x)sec2(x)=8sin2(x)
8sin2(x)cos2(x)=1
This gives us
2sin2(2x)=1sin(2x)=±12
2x=nπ2+π4
x=nπ4+π8

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