Solve it 4sin4theta+1=root5?

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Answer 1 · 2017-10-29T08:32:56

#theta# = #(npi)/2# +#pi/40#
#theta# = #(npi)/2#+#(9pi)/40#
for all real integers n

Rearanging;
#sin4theta = (5^(1/2) -1)/4#
#sin4theta = sin(arcsin( (5^(1/2) -1)/4 )) #
#sin4theta = sin(pi/10)#

Now we can consider the general solution to sin#x# = #sina#

The solution to this is #x# = #2npi + a# and #x# = #2npi + pi - a# for all integers n

so hence #4theta = 2npi + pi/10 and 4theta = 2npi + (9pi)/10 #

Hence dividing by 4, yields;

#theta# = #(npi)/2# +#pi/40#
#theta# = #(npi)/2#+#(9pi)/40#

for all real integers n