Question

1) sin 2x + cos 2x= 0 2) sin 4x - cos 2x = 0

Answer 1

$\sin2x+\cos 2x=0|:\cos 2x \\ tg2x+1=0 \\ tg2x=-1 \\ 2x=- \frac{ \pi }{4} + \pi n, n \in Z \\ x=- \frac{ \pi }{8} + \frac{ \pi n}{2} , n \in Z$

$\sin 4x-\cos 2x=0 \\ 2\sin 2x\cos 2x-\cos 2x=0 \\ \cos 2x(2\sin2x-1)=0 \\ \left[\begin{array}{ccc}\cos 2x=0\\\sin 2x= \frac{1}{2} \end{array}\right\to \left[\begin{array}{ccc}2x= \frac{ \pi }{2}+ \pi n, n \in Z\\2x=(-1)^k\cdot \frac{ \pi }{3}+ \pi k, k \in Z \end{array}\right \\ \\ \left[\begin{array}{ccc}x_1= \frac{ \pi }{4} + \frac{ \pi n}{2} , n \in Z\\x_2=(-1)^k\cdot \frac{ \pi }{6} + \frac{ \pi k}{2},k \in Z\end{array}\right$

Answer 2

Let's solve each equation separately:

1. sin 2x + cos 2x = 0:

To solve this equation, we can use a trigonometric identity to rewrite it in terms of a single trigonometric function. The identity we'll use is:

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

Replacing sin(2x) and cos(2x) in the equation, we have:

2sin(x)cos(x) + cos^2(x) - sin^2(x) = 0

Rearranging the terms:

cos^2(x) + 2sin(x)cos(x) - sin^2(x) = 0

Now, we can use another identity to simplify further:

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

Substituting in the equation:

cos(2x) + 2sin(x)cos(x) = 0

Factoring out a common factor:

cos(2x) + 2sin(x)cos(x) = cos(2x) + sin(2x) = 0

Now we have:

cos(2x) + sin(2x) = 0

Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation as:

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

Dividing both sides by cos(2x) (assuming cos(2x) is not equal to zero), we get:

tan(2x) = -1

Taking the inverse tangent (arctan) of both sides, we have:

2x = arctan(-1)

Simplifying further:

2x = -π/4 + nπ (where n is an integer)

Dividing by 2:

x = -π/8 + nπ/2 (where n is an integer)

So the general solution to the equation sin 2x + cos 2x = 0 is:

x = -π/8 + nπ/2 (where n is an integer)

2. sin 4x - cos 2x = 0:

Using similar steps as above, we can rewrite the equation as:

2sin(2x)cos(2x) - cos^2(2x) + sin^2(2x) = 0

Rearranging the terms:

sin^2(2x) + 2sin(2x)cos(2x) - cos^2(2x) = 0

Now, let's use another trigonometric identity:

sin^2(x) + cos^2(x) = 1

Substituting in the equation:

2sin^2(2x) + 2sin(2x)cos(2x) - 2cos^2(2x) = 0

Dividing the equation by 2:

sin^2(2x) + sin(2x)cos(2x) - cos^2(2x) = 0

Factoring the equation:

(sin(2x) - cos(2x))(sin(2x) + cos(2x)) = 0

Since sin(2x) + cos(2x) = 0 (as given in the equation), we have:

sin(2x) - cos(2x) = 0

Now we can solve this equation as we did in the previous example:

tan(2x) = 1

Taking the inverse tangent:

2x = π/4 + nπ (where