How do you find the general solutions for #sinx = cos2x#?

2 Answers
Muhammad Arifur Rahman
Sep 18, 2015

#x=30^{\circ}, 270^{\circ}# #[0^0 \leq x \leq \360^0]#

Explanation:

We know,

#cos2x=cos^2x-sin^2x=1-2sin^2x#

So, let's solve the equation now,

#sinx=cos2x=1-2sin^2x#
#\rightarrow 2sin^2x+sinx-1=0#
#\rightarrow 2sin^2x+2sinx-sinx-1=0#
#\rightarrow 2sinx(sinx+1)-1(sinx+1)=0#
#\rightarrow (2sinx-1)(sinx+1)=0#

Now,

#2sinx-1=0#
#\rightarrow sinx=\frac{1}{2}#
#\rightarrow x=sin^{-1}(##\frac{1}{2}#)
#\rightarrow x=30^{\circ}#

And, #sinx+1=0#
#\rightarrow x=sin^{-1}(-1)##= 270^{\circ}#

As we just need the general solutions, we should take only this two values as the general solutions .
Answer : #30^0, 270^0#
That's it!

Nghi N.
Sep 18, 2015

Solve sin x = cos 2x

Explanation:

Apply trig identity: #cos 2x = 1 - 2sin^2 x#
#sin x = 1 - 2sin^2 x#. Solve the quadratic equation:
#2sin^2 x + sin x - 1 = 0#
Since (a - b + c = 0), use Shortcut. Two real roots: sin x = -1 and #sin x = -c/a = 1/2#.

#sin x = 1/2# --> x = 30 deg and x = 150 deg #(pi/6 and (5pi)/6)#
sin x = -1 --> x = 270 deg #((3pi)/2)#
General solutions:
x = 30 + k360 deg
x = 150 + k360
x = 270 + k360.

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