Question

help guys will medal and fan:
tan(x)=cos(x) solve the equ.?

Answer 1

tan(x) = sin(x)/cos(x)
sin(x)/cos(x) = cos(x) * cos(x) sin(x) = cos^2(x)
cos^2(x) = 1 - sin^2(x)
sin(x) = 1- sin^2(x)
sin^2(x) + sin(x) - 1 = 0

Answer 2

@TrojenPoem thxxx man

Answer 3

a = 1 b = 1 c = -1
-b (+-) sqrt((b)^2 - 4ac)/ 2a
-1 (+-)sqrt(1 - 4 * 1 * -1) / 2
-1 (+-)sqrt(5)/2
sinx = -1+sqrt(5)/2 and sinx = 1-sqrt(5)/2

Answer 4

yw

Answer 5

ive got Another ques part b??

Answer 6

@TrojanPoem ?

Answer 7

which one ?

Answer 8

part b from above

Answer 9

@TrojanPoem any idea?

Answer 10

Thinking

Answer 11

okay you we need to solve the angle x to satisfy the equ man it literary fluttered me 😂🙈

Answer 12

you know*

Answer 13

I am thinking about dividing the whole eqn by 2 and then multiply sqrt(2)/2 in
so I will get sinxcos(45) + cosx sin(45) which is the sum of two angles sin(x+45)
now still thinking..

Answer 14

correct but still not reachable try harder and ill do too

Answer 15

sin2x + 2 = 4sin(x + 45) trying.

Answer 16

then

Answer 17

No clue. @phi

Answer 18

@TrojenPoem ahhh ok are you good with integrals?

Answer 19

@TrojanPoem

Answer 20

s c + 1 = sqrt(2) (s+c)
square both sides
(s c)^2 + 2 sc + 1 = 2(s^2 + c^2 + 2s c)
s^2 + c^2 = 1, so

(sc)^2 + 2sc + 1 = 2 + 4 sc
(sc)^2 - 4sc -1 = 0

let sc= x, and solve for x
x= 1 +/- sqr(2)
the 1+ sqr(2) is larger than 1 , so we would get imaginary angles

solve
\[ \sin x \cos x = 1 - \sqrt{2} \]

Answer 21

sin(2x) = 2 - 2 sqrt(2)
2x = asin(2 (1 - sqrt(2))

Answer 22

\[sinx.cosx+1=\sqrt{2}(sinx+cosx)\]\[(sinx.cosx+1)^{2}=2(sinx+cosx)^{2}\]\[\sin^2x.\cos^2x+2.sinx.cosx+1=2(\sin^2x+\cos^2x+2.sinx.cosx)\]\[\sin^2x.\cos^2x+2.sinx.cosx+1=2(1+2.sinx.cosx)\]\[\sin^2x.\cos^2x+1=2+2.sinx.cosx\]\[\sin^2x.\cos^2x-2.sinx.cosx-1=0\]\[sinx.cosx=t\]\[t^2-2t-1=0\]\[t=\frac{2 \pm \sqrt{4-(4.1.(-1))}}{2}=\frac{2 \pm \sqrt{8}}{2}=\frac{2 \pm2\sqrt{2}}{2}\]\[t=1 \pm \sqrt{2}\]\[sinx.cosx=1 \pm \sqrt{2}\]\[\sin(2x)=2 \pm 2\sqrt{2}\]\[\sin(2x)\neq2+2\sqrt{2}\]\[\sin(2x)=2-2\sqrt{2}\]\[2x=\sin^{-1}(2(1-\sqrt{2}))\]\[x=\frac{1}{2}\sin^{-1}(2(1-\sqrt{2}))\]

Answer 23

looks familiar

Answer 24

@Nishant_Garg I love you mannn :p thanks alot it fluttered me