Sqrt(1-3x)=x+3
[sqrt(1-3x)]^2=(x+3)^2
1-3x=(x+3)(x+3)
1-3x=x^2+6x+9
-3x=x^2+6x+8
0=x^2+9x+8
The answers are -1 and -8 BUT, we have to plug them back into the original equation to make sure we don't get a negative under the square root sign.
After doing this, we realize that only -1 works, so the answer is x=-1
Sorry that this took forever to answer. I was thinking of a good way to explain this, and if you need any further explanation, message me:)
Best wishes:)
1 ) cot x * sin x = cos x
(cos x / sin x) * sin x = cos x
cos x = cos x
Answer: B ) cot x = cos x / sin x
2 ) ( sin² x + cos² x ) / cos x = sec x
1/cos x = sec x
sec x = sec x
Answer: C ) cos² x + sin² x = 1
Esmeralda I think it's wrong cuz it’s like the question but that's added a parentheses so I think that's wrong
The answer is 22 , you’re welcome!
