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Question

How many values of xϵ[0,2π] satisfies the equation sin 2x + 5 sin x + 1 + 5 cos x = 0?


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Solution

When we have trigonometric equations involving (sinx±cosx) and sin x cos x or sin 2x, we can solve it

by the substitution sinx±cosx=t.
If we observe the expression given, we can re-write it as

1+ sin 2x + 5(sin x + cos x)
This is an expression involving sin x + cos x and sin 2x. so we will go for the substitution sin x+ cos x = t.

It is important to know that sin x + cos x and sin 2x are connected by the relation

(sinx+cosx)2 =1+ sin 2x. [we also have (sinxcosx)2=1-sin 2x]
1 + sin 2x + 5(sin x + cos x) = t2 + 5t.

t2 + 5t = 0

t=0 or t+5=0

sin x + cos x = 0 or sin x + cos x + 5 = 0

sin x + cos x + 5 can't be zero, because minimum value of sin x + cos x is -2

sin x + cos x = 0

tan x = -1

= tan π4

x=nπ - π4 , when we put n = 0,1,2,3, ........ we get,

x = π4,3π4,7π4,11π4,15π4 ...... out of which 3π4 , 7π4 , belongs to [0,2π]. Therefore two values


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