Question

Solve the following equation for 0° ≤ θ < 360°. Use the "^" key on the keyboard to indicate an exponent. For example, sin2x would be typed as sin^2x. Be sure to show all your work.

2sin2x - cos2 x - 2 = 0

Answer 1

This is the guy aboves work, I just wrote it out so you can copy and paste easier.

2sin^2x-cos^2x-2=0

sin^2x+cos^2x=1

cos^2x=1-sin^2x

2sin^2x-(1-sin^2x)-2=0

2sin^2x -1 +sin^2x-2=0

3sin^2x-3=0

3sin^2x=3

sin^2x=1

sin x =(sqr root)1

sin x= + or - 1

sin x =1 sin x = -1

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

x=90+n360 x=270 + 360n

x=90, 270 (wrote down so it stands out more)

Answer 2

`\boxed{\boxed{x=90^(\circ),270^(\circ)}}`

Solution-

The given equation-

`\Rightarrow 2\sin^2 x-\cos^2 x-2=0`

As

`\sin^2 x+\cos^2 x=1\ \ \Rightarrow \cos^2 x=1-\sin^2 x`

Putting this,

`\Rightarrow 2\sin^2 x-(1-\sin^2 x)-2=0`

`\Rightarrow 2\sin^2 x-1+\sin^2 x-2=0`

`\Rightarrow 3\sin^2 x-3=0`

`\Rightarrow 3\sin^2 x=3`

`\Rightarrow \sin^2 x=1`

`\Rightarrow \sin x=\sqrt1`

`\Rightarrow \sin x=\pm 1`

`\Rightarrow \sin x=1,\ \sin x=-1`

`\Rightarrow x=\sin^(-1)(1),\ x=\sin^(-1)(-1)`

`\Rightarrow x=90^(\circ)+n360^(\circ),\ x=270^(\circ)+n360^(\circ)`

Where n=0, 1, 2, ......

As given 0° ≤ x < 360°, so putting n = 0

`\Rightarrow x=90^(\circ),\ x=270^(\circ)`