Trig question, what have I don't wrong? Solve (tan(2x)-tan(x))*(cos(3x)+cos(x))+2*cos(x)=2?

Jade

2021-01-02 3:23 pm

Solve (tan(2x)-tan(x))(cos(3x)+cos(x))+2cos(x)=2, 0 <= x <= pi
(sinx/(cosx cos2x))(2cos2x cosx)+2cosx=2 <---- (*)
2sinx+2cosx=2
sin^2(x)+cos^2(x)+2cosxsinx =1
2sinxcosx=0
sin2x=0
2x=0, pi, 2pi
x = 0, pi/2, pi

x=pi/2 is rejected because tan(pi/2) is undefined.
how about x=pi? if I sub x=pi into the equation, I get -2 instead of 2.
What have I don't wrong?

(*) tan2x-tanx
=(2tanx/(1-tan^2(x)))-tanx
=(tanx(1+tan^2(x)))/(1-tan^2(x))
=tanxsec^2(x)/(1-tan^2(x))
=tanx/(cos^2(x)-sin^2(x))
=sinx/cosxcos2x

回答 (1)

[匿名]

2021-01-02 4:32 pm

✔ 最佳答案

Question：
Solve [tan(2x) - tan(x)][cos(3x) + cos(x)] + 2cos(x) = 2, 0 ≤ x ≤ π
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Solution：
[tan(2x) - tan(x)][cos(3x) + cos(x)] + 2cos(x) = 2, 0 ≤ x ≤ π
...
sin x + cos x = 1
(sin x + cos x)² = 1 ...... [ It also implies (sin x + cos x)² = (-1)² ]
...
x = 0, π, π/2 ( rejected )

[ x = 0 is the solution of sin x + cos x = 1. ]
[ x = π is the solution of sin x + cos x = -1. ]

∵ [tan(2x) - tan(x)][cos(3x) + cos(x)] + 2cos(x) = 2 ⇒ sin x + cos x = 1
∴ x = π should be rejected.

Alternate method：
[tan(2x) - tan(x)][cos(3x) + cos(x)] + 2cos(x) = 2, 0 ≤ x ≤ π
...
sin x + cos x = 1
sin x + sin(π/2 - x) = 1
2 sin(π/4) cos(x - π/4) = 1
cos(x - π/4) = 1/√2
x - π/4 = -π/4, π/4
x = 0, π/2 ( rejected )

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