Solve 4 sinx + 9 cosx=0 for 0° ​

4 sin(x) + 9 cos(x) = 0

4 sin(x) = -9 cos(x)

tan(x) = -9/4

x = arctan(-9/4) + nπ … … … (in radians)

or

x = arctan(-9/4) + 180n ° … … … (in degrees)

where n is any integer.

I'm guessing you're solving for x over some domain, probably 0° ≤ x < 360°. In that case, you would have two solutions for n = 1 and n = 2 of

x ≈ 113.96°   and   x ≈ 293.96°

A-b=21
5a-2b=18

Set up the first equation so that it equals a or b. This method is using substitution to help solve the problem. So, if you add the inverse of a or b.. So let's say..

a-b(+b)=21+b

So, obviously the variables cancel out and you make sure you follow the rule of what you do to one side, you do to the other. Now you're left with

a=21+b

Now plug in this equation above into your second equation, and solve. :)

So,

5(21+b)-2b=18

Distribute
21 x 5 = 105
5 x b = 5b

Now,
105 + 5b -2b=18
105 +3b=18

105-105+3b=18-105
3b=-87
b= -29

Now plug in b into the first (original) equation:

a-(-29)=21

Since 2 negatives equal a positive, then it is written as follows:

a+29=21

Subtract.

21-29=-8

a=-8

So to check you answer plug in the values of the first equation of a-b=21. So, (-8)-(-29)=21. Again 2 negatives equal a positive so it become -8+29. So this equals a positive 21. :) Best wishes!