C2 Chapter 10 - Trigonometric Identities Question

Solve 2$sinxcosx+cos^2x= 0$

for 0 < x < 360

What are the solutions?

i've solved $cosx = 0$ as 90 and 270 degrees?

i cant solve the other one :?

Move the cos^2x over to the other side. Divide by cosx. Then you should know what to do.

You will lose some solutions if you do it in this way. You also need to solve it for $cosx=0$

Solve 2sinxcosx + cos^2x

for 0 < x < 360

What are the solutions?


I'd start off by using sinx/cosx= tanx
move cos^2x to the other side.
you get 2sinxcosx=-cos^2x
divide through by cos x ---> 2sinx= -1
sinx= -1/2
solution one---> shiftsin(-1/2)= -30 degrees
solution two ---> sine is positive in Q2: 180 degrees- (-30)= 210 degrees
other solution(s)---> sine graph has period of 360 degrees (.ie. it repeats every 360 degrees) so +/- 360 to both solutions to find other solutions satisfying range. we can only add 360 degrees to the -30 and not to the 210 degrees. Nor can we subtract 360 degrees from either solutions as the values we would obtain would be out of range. Therefore, the solutions within the range of 0<x<360 are 210 and 330 degrees.

Hopefully I've got this correct

You have got two correct solutions.

For the other one, $2sinx+cosx=0 \implies 2sinx=-cosx$

Now divide both sides by $cosx$ to get an equation involving only $tanx$.

okay..

im at tanx + sinx + cosx = 0

do you square everything and then end up with...

$tan^2x + sin^2x + cos^2x = 0[br][br]tan^2x + 1 = 0[br][br]tan^2x = -1$

root and inverse tan

x = inverse tan (1)

quadrant rule and x = 45 + 225 along with 90 + 270

is this right?

$2sinxcosx + cos^2x = 0 \implies cosx(2sinx+cosx)=0$

Now we get two equation, $cosx=0 \text{ and } 2sinx+cosx=0$, solve both the equations to find the solutions.

$2sinxcosx + cos^2x = 0 \implies cosx(2sinx+cosx)=0$

Now we get two equation, $cosx=0 \text{ and } 2sinx+cosx=0$, solve both the equations to find the solutions.

how did you get the bracket bit

is it because cosx is common in both?

is it because cosx is common in both?

I just pulled cosx as a common factor.

Yes. It's always best to try get something = 0 into a product of factors = 0 if possible as then you can tackle each bracket independently being = 0 (like when solving a quadratic)