+564 Solve the following equations given that 0 ≤ x < 2pi
$$cos^2x - sin^2x = -1$$
$${\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)} = {\mathtt{0}}$$
+129899 For the first one, we have
cos2x - sin2x = -1
Note that sin2x = 1 - cos2x So we have
cos2x - (1 - cos2x ) = -1
2cos2x - 1 = -1 add 1 to both sides
2cos2x = 0 divide by 2 on both sides
cos2x = 0 take the suare root of both sides
cosx = 0 So x = pi/2 and x =3pi/2
For the second one, we have
1 + √(3) tan(2x) = 0 Subtract 1 from both sides
√(3) tan(2x) =-1 Divide both sides by √(3)
tan(2x) = -1/√(3)
Note that tan(x) = -1/√(3) at 150 degrees and at 330 degrees
So tan(2x) = -1/√(3) at 75 degrees and at 165 degrees
And converting these to rads, we have x = (5pi)/12 and x = (11pi)/12
+129899 For the first one, we have
cos2x - sin2x = -1
Note that sin2x = 1 - cos2x So we have
cos2x - (1 - cos2x ) = -1
2cos2x - 1 = -1 add 1 to both sides
2cos2x = 0 divide by 2 on both sides
cos2x = 0 take the suare root of both sides
cosx = 0 So x = pi/2 and x =3pi/2
For the second one, we have
1 + √(3) tan(2x) = 0 Subtract 1 from both sides
√(3) tan(2x) =-1 Divide both sides by √(3)
tan(2x) = -1/√(3)
Note that tan(x) = -1/√(3) at 150 degrees and at 330 degrees
So tan(2x) = -1/√(3) at 75 degrees and at 165 degrees
And converting these to rads, we have x = (5pi)/12 and x = (11pi)/12
