How do you solve for x in this equation? ?

Question

(cos^2)(x)-(sin^2)(x)=sinx

Answer 1

Use C^2x = 1 - S^2x

(1 - S^2x) - S^2x = Sx

1- 2S^2x = Sx

or S^2x + 0.5 Sx - 0.5 = 0

Let u = Sx

u^2 + 0.5 u - 0.5 = 0

(u - 0.5) (u+1) = 0

So U = Sx = 1/2 or -1

x = pi/6 (30 deg), 5pi/6(150 deg) for Sx=1/2

x = 3pi/2 for Sx=-1

Answer 2

NOTE: cos^2(x) = cos²(x) .  assume your web browser do show this correctly

cos²(x) - sin²(x) = sin(x)

>> trig ID: cos²(x) + sin²(x) = 1   →   cos²(x) = 1 - sin²(x)

[1 - sin²(x)] - sin²(x) = sin(x)

-2sin²(x) = sin(x) - 1

0 = 2sin²(x) + sin(x) - 1

factorize:

[2sin(x) - 1][sin(x) + 1] = 0

find the "zero's"

2sin(x) - 1 = 0     OR     sin(x) + 1 = 0

the first part:

2sin(x) - 1 = 0

2sin(x) = 1

sin(x) = 1/2

x = sin^-1(1/2) = arcsin(1/2)

note that

sin(x) = opposite/hypotenuse = 1/2, then

arcsin(opposite/hypotenuse) = arcsin(1/2)

if you find y=1/2 on a r=2 circle, then you find that

x = 30°   {degrees}

x = pi / 6   {radians}

second part:

sin(x) + 1 = 0

sin(x) = -1

sin(x) = -1/1

on y= -1 on a r=1 circle

x = -90° {degrees}

x = - pi / 2  {radians}

ANSWER

cos²(x) - sin²(x) = sin(x)

x = 30°   OR   x = -90°      {degrees}

x = pi/6   OR   x = - pi / 2     {radians}

------

I hope this helps

love

@}'-,-'-,-

Answer 3

sinx= 1/2 or -1 then x= 30, 270

Answer 4

Ahh!!!! My head hurts just looking at it!!  

Answer 5

(cos^2)(x)-(sin^2)(x)=sinx

1-sin^2(x)-sin^2(x)=sin(x) [sin^2(x)+cos^2(x)=1]

1-2sin^2(x)=sin(x)

1=sin(x)-2sin^2(x)

1=sinx {1-2sin(x)}  By taking common

sin(x)=1  x=90 degree

or

1-2sin(x)=1

2sin(x)=0

sin(x)=0

Rate me if u find my ans

x=0 degree

Ans-0 degree 0r 90 degree

Rate me if u find my ans helpful.

:)))

Answer 6

there are rules that should be in your book about these probles. I forgot the rules and I am sorry. You will find them easy but again i am sorry I forgot the rules.. Maybe something like

sin^2=1-cos^2x..hope i helped!