Trig help! (1 Viewer)

nightweaver066 · Feb 15, 2012

JasonNg said:
Repped you guys! Can you guys show me how to do this last one? Solve for x,

6 tan^2x - 4sin^2x = 1

Thanks!

$6tan^2x - 4sin^2x = 1$

$6\frac{sin^2x}{cos^2x} - 4sin^2x - 1 = 0$

$6sin^2x - 4sin^2xcos^2x - cos^2x = 0$

$6sin^2x - 4sin^2x(1 - sin^2x) - (1 - sin^2x) = 0$

$6sin^2x - 4sin^2x + 4sin^4x - 1 + sin^2x = 0$

$4sin^4x + 3 sin^2x - 1 = 0$

$4sin^4x + 4sin^2x - sin^2x -1 = 0$

$(4sin^2x - 1)(sin^2x + 1) = 0$

$sin^2x = \frac{1}{4},\; sin^2x = -1$

$$As $ 0 \leq sin^2x \leq 1$

$sin^2x = \frac{1}{4}$

$sinx = \pm \frac{1}{2}$

$x = 30, 150, 210, 330$

nightweaver066 · Feb 15, 2012

Carrotsticks said:
Haha nightweaver, how did you manage to get the upside down question mark?

I have no clue what you're talking about but i'm curious myself.

Carrotsticks · Feb 15, 2012

nightweaver066 said:
I have no clue what you're talking about but i'm curious myself.

I could have sworn I saw an upside down question mark in your original working out, before you deleted it.

nightweaver066 · Feb 15, 2012

Carrotsticks said:
I could have sworn I saw an upside down question mark in your original working out, before you deleted it.

How embarrassing.. The original working out was up for less than 10 seconds. Thought i could get away with it.

Carrotsticks · Feb 15, 2012

nightweaver066 said:
Thought i could get away with it.

Never.

Timothy6340 · Feb 16, 2012

double angles is only 3unit i think

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Trig help! (1 Viewer)

nightweaver066

Well-Known Member

nightweaver066

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Carrotsticks

Retired

nightweaver066

Well-Known Member

Carrotsticks

Retired

Timothy6340

Member

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