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-/+ Root Solutions

10cos(xa)=2\sqrt{10}cos(x-a)=2

The mark scheme lists the only solutions when 2 is divided by positive root 10

Why wouldn't the solutions where 2 is divided by negative root 10 be acceptable?
Reply 1
Original post by pleasedtobeatyou
10cos(xa)=2\sqrt{10}cos(x-a)=2

The mark scheme lists the only solutions when 2 is divided by positive root 10

Why wouldn't the solutions where 2 is divided by negative root 10 be acceptable?


4=2±2 \sqrt4 = 2 \not= \pm 2

The value obtained from the square root is always taken as positive.
maybe there is some limitation on x, like 0xπ0\leq x\leq \pi
This is from Core 4 but in some Core 3 questions for example,

tan2x=6tan^2x = 6

There's a need to find the solutions for -root6 and +root6
Original post by njl94
maybe there is some limitation on x, like 0xπ0\leq x\leq \pi


Nope, the solutions for the negative root would also be acceptable
Reply 5
Original post by pleasedtobeatyou
This is from Core 4 but in some Core 3 questions for example,

tan2x=6tan^2x = 6

There's a need to find the solutions for -root6 and +root6


tan2x=6    tanx=±6 \tan^2 x = 6 \implies \tan x = \pm \sqrt{6}

It is the same as x2=4    x=±4=±2 x^2 = 4 \implies x = \pm \sqrt4 = \pm 2

But x=4=2±2 x = \sqrt4 = 2 \not= \pm \sqrt2

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