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Mathematics 19 Online
OpenStudy (anonymous):

secx*cscx= 2secx how can I find the solution?

OpenStudy (amistre64):

essentially: ab = 2a

OpenStudy (amistre64):

ab = 2a ab-2a = 0 a (b-2) = 0

OpenStudy (anonymous):

\[\sec x \csc x -2\sec x=0\] \[\sec x \left( \csc x-2 \right)=0\] now you can solve it.

OpenStudy (amistre64):

or, since secx is never zero ab = 2a b = 2 1/b = 1/2

OpenStudy (anonymous):

Alrighty,I have one solution but i don't know if there is others

OpenStudy (amistre64):

well, pi/6 and 5pi/6 seem to work for me

OpenStudy (anonymous):

\[1/\cos(x)•1/\sin(x)=2/\cos(x)\] Multiplying by cos(x), \[\cos(x)•1/\cos(x)•1/\sin(x)=\cos(x)•2•1/\cos(x)\] \[1/\sin(x)=2\] Multiplying by sin(x), \[\sin(x)•1/\sin(x)=2•\sin(x)\] \[1=2\sin(x)\] \[\sin(x)=1/2\] \[π/6+2πn, 5\pi/6+2πn\] (or, if you need it in degrees, 30º+n360º and 150º+n360º)

OpenStudy (amistre64):

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OpenStudy (anonymous):

would 11pi/6 be one too?

OpenStudy (amistre64):

no

OpenStudy (anonymous):

oh okay

OpenStudy (amistre64):

but essentially, its all turns that get you into pi/6 and 5pi/6

OpenStudy (amistre64):

1,13,25,... +12 each time 5,17,29,... +12 each time 11 aint there

OpenStudy (anonymous):

because sin x is positive ,it has to be in first and second quadrant only.

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