Question

I need help please, I am not sure how to approach it: sinx-cosx=0 (Find all solutions)

Answer 1

Am I supposed to do the quotient identity?

Answer 2

\(\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)
\\ \quad \\
{\color{brown}{ sin(\theta)=\pm \sqrt{1-cos^2(\theta)}}}\qquad thus
\\ \quad \\ \quad \\
sin(x)-cos(x)=0\implies \sqrt{1-cos^2(x)}-cos(x)=0
\\ \quad \\
\sqrt{1-cos^2(x)}=cos(x)\implies 1-cos^2(x)=[cos(x)]^2
\\ \quad \\
1-cos^2(x)=cos^2(x)\implies 1=2cos^2(x)\implies \cfrac{1}{2}
\\ \quad \\
\pm \sqrt{\cfrac{1}{2}}=cos(x)\)

then take the arcCos() to each side to get "x"

Answer 3

hmm anyhow missing a bit \(\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)
\\ \quad \\
{\color{brown}{ sin(\theta)=\pm \sqrt{1-cos^2(\theta)}}}\qquad thus
\\ \quad \\ \quad \\
sin(x)-cos(x)=0\implies \sqrt{1-cos^2(x)}-cos(x)=0
\\ \quad \\
\sqrt{1-cos^2(x)}=cos(x)\implies 1-cos^2(x)=[cos(x)]^2
\\ \quad \\
1-cos^2(x)=cos^2(x)\implies 1=2cos^2(x)\implies \cfrac{1}{2}=cos^2(x)
\\ \quad \\
\pm \sqrt{\cfrac{1}{2}}=cos(x)\)

Answer 4

are you square rooting it because the sine has an exponent of 2?

Answer 5

hm,m?

Answer 6

what do you mean?

Answer 7

I don't understand why the sine and cosine are squared