trig identities

[abstract field]

help i tried i used 1= sin^2x + cos^2
then got
5sinx(sinx+cosx)=3(sin^2x+cos^2x)

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[jolly valley]

Did you expand the brackets?

[abstract field]

yeah

it got me no where

i got

5sin^2x + cosx = 3sin^2+ 3cos^2

[abstract field]

do u k

[jolly valley]

I think i might know

[abstract field]

can u tell me

i k the answer

but don't know how to get it

[jolly valley]

So you get 5sin^2(x) + 5sinxcosx = 3 so
5sinxcosx = 3-5sin^2(x)

Then squaring both sides gives
25sin^2(x)cos^2(x) = 9 - 30sin^2(x) + 25sin^4(x)

Then you can substitute cos^2(x) = 1-sin^2(x) on the left hand side to get
25sin^2(x) [1-sin^2(x)] = 9- 30sin^2(x) + 25sin^4(x)

Then you expand the left hand side

To get 25sin^2(x) - 25sin^4(x) = 9 - 30sin^2(x) + 25sin^4(x)

Then rearranging gives the quadratic
0 = 9 - 55sin^2(x) + 50sin^4(x)

Wait how many marks is this

[abstract field]

idfk

my maths teacher is crazy

[jolly valley]

Cuz this feels like i'm gonna do like 10 marks worth

[abstract field]

he gives us random shit to complete

i don't think thats how u doi it dear

[jolly valley]

No this method works

Cuz it factorises to 0 = (10sin^2(x) - 9)(5sin^2(x) - 1)

So sin^2(x) = 9/10 or sin^2(x) = 1/5

Then its just sinx = ±3/√10 or sinx = ±1/√5

And inverse them

But theres probably a quicker way

[abstract field]

see the answers give me

26.6

[jolly valley]

Ye thats what i get

For one of the solutions

[abstract field]

ok how in the wolrd to i write that donw idk

[jolly valley]

Probably ask your teacher cuz theres probably a faster solution

[abstract field]

its due tmmr man he ain't gonna be happy

oh well

[jolly valley]

I mean its a pretty hard question