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Stuck on C3 trig question

SOlve in the interval 0 to 2pi
cos2x=tan2x

I tried to get it all in terms of sinx but have gotten to (1-2sin^2 x)=2sinxcosx

Which identies do I use?
Original post by bobbricks
SOlve in the interval 0 to 2pi
cos2x=tan2x

I tried to get it all in terms of sinx but have gotten to (1-2sin^2 x)=2sinxcosx

Which identies do I use?


Why have you changed to x

Leave the question in 2x until the final answer line
Reply 2
Avatar for davros
davros
16
Original post by bobbricks
SOlve in the interval 0 to 2pi
cos2x=tan2x

I tried to get it all in terms of sinx but have gotten to (1-2sin^2 x)=2sinxcosx

Which identies do I use?


There are probably several ways of attacking this, but I would just multiply by cos 2x to clear fractions so you've got

cos2(2x)=sin2xcos^2(2x) = sin2x

Use a standard identity on the LHS and you have a quadratic equation in the new variable sin2x to solve.
Reply 3
Avatar for bobbricks
bobbricks
OP
19
Can you use sin²(2x)+cos²(2x)=1 ?
Original post by bobbricks
Can you use sin²(2x)+cos²(2x)=1 ?


Yes

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