How do I solve for the value of x in degrees if √2sinx-1=0?

1 Answer
Jack H.
Feb 21, 2018

# x=135°+360°*n# or #x=360°*n+45°#

Explanation:

First, let's solve it in radians

#sqrt2sinx-1=0#
#therefore sinx=1/sqrt2=sqrt2/2#
If you look at Special Ratio, #sin45°# is #sqrt2/2#
proof)
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SOH CAH TOA, #sin45°=(Opposite)/(Hypotenuse)#
#therefore 1/sqrt2=sqrt2/2#

But this was the only case when #0°<=x<=90°(0<=x<=pi/2)#

Now we need to look on real number range

Since #sin# has period of #2pi#, (#f(?)=f(?+2pi)#)

#thereforex=360°*n+45°(2npi+pi/4)#

But we know that #sin# function is positive @ 2nd quadrant.#(-, +)# Which means,
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We need to find out the #sinx=sqrt2/2# for 2nd quadrant #(90°<=x<=180°(pi/2<=x<=pi))#
#x=(3pi)/4#
But #sinx# has period of #2pi#,

#therefore x=(3pi)/4+2npi#

If you write them in degrees,

#x=(3*180°)/4+360°*n=135°+360°*n#

#therefore x=135°+360°*n#

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