Question

Knowing that \(2\sin x + \cos x = 1\), find \(\tan x\).

Answer 1

x=0, the first guess :)

Answer 2

Are there any more solutions?

Answer 3

2sinx+cosx=1
divided by cosx to both sides giving us
2tanx + 1 = secx
squares to both sides :
4tan^2 x + 4tanx + 1 = sec^2 x
4tan^2 x + 4tanx + 1 = 1 + tan^2 x
3tan^2 x + 4tanx = 0
tanx(3tanx + 4) = 0

Answer 4

take the zeroes of all factors

Answer 5

btw, u should give intervals for x

Answer 6

\[2sinx=1-cosx\]
after applying half-angle formulae,
\[2*2\sin \frac{x}{2}\cos \frac{x}{2}=2\sin^{2} \frac{x}{2}\]
\[2\cos \frac{x}{2}=\sin \frac{x}{2}\]
\[\tan \frac{x}{2}=2\]
apply this for tanx.
\[\tan(\frac{x}{2}+ \frac{x}{2}) = \frac {\tan(\frac {x}{2})+\tan(\frac {x}{2})}{1-\tan(\frac {x}{2})\tan(\frac {x}{2})}\]

Answer 7

one more alternate method, for \(a\sin x+b\cos x=c\),
divide both sides by \(\sqrt{a^2+b^2}\)