Answer :
Answer-
[tex]\boxed{\boxed{x=90^{\circ},270^{\circ}}}[/tex]
Solution-
The given equation-
[tex]\Rightarrow 2\sin^2 x-\cos^2 x-2=0[/tex]
As
[tex]\sin^2 x+\cos^2 x=1\ \ \Rightarrow \cos^2 x=1-\sin^2 x[/tex]
Putting this,
[tex]\Rightarrow 2\sin^2 x-(1-\sin^2 x)-2=0[/tex]
[tex]\Rightarrow 2\sin^2 x-1+\sin^2 x-2=0[/tex]
[tex]\Rightarrow 3\sin^2 x-3=0[/tex]
[tex]\Rightarrow 3\sin^2 x=3[/tex]
[tex]\Rightarrow \sin^2 x=1[/tex]
[tex]\Rightarrow \sin x=\sqrt1[/tex]
[tex]\Rightarrow \sin x=\pm 1[/tex]
[tex]\Rightarrow \sin x=1,\ \sin x=-1[/tex]
[tex]\Rightarrow x=\sin^{-1}(1),\ x=\sin^{-1}(-1)[/tex]
[tex]\Rightarrow x=90^{\circ}+n360^{\circ},\ x=270^{\circ}+n360^{\circ}[/tex]
Where n=0, 1, 2, ......
As given 0° ≤ x < 360°, so putting n = 0
[tex]\Rightarrow x=90^{\circ},\ x=270^{\circ}[/tex]
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)