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

Answer-

$\boxed{\boxed{x=90^{\circ},270^{\circ}}}$

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

$\Rightarrow x=90^{\circ},\ x=270^{\circ}$