# Trigonometric equation sin^4x+cos^4x=sin2x - 1/2

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

### Trigonometric equation sin^4x+cos^4x=sin2x - 1/2

Hi!I need some help...
sin^4x+cos^4x=sin2x - 1/2

10x
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### Re: Trigonometric equation sin^4x+cos^4x=sin2x - 1/2

$$sin^{4 }x + cos^{4 }x$$= sin(2x) -$$\frac{1}{2}$$

$$( sin^{2 }x + cos^{2 }x )^{2 }$$ -2$$sin^{2 }x. cos^{2 }x$$ =2sinxcosx -$$\frac{1}{2}$$

2$$(sinx.cosx)^{2 }$$ +2(sinx.cosx) -$$\frac{3}{2}$$=0
a=sinx.cosx
4$$a^{2 }$$+4a-3 =0 $$\Rightarrow$$ $$a_{1 }$$= $$\frac{1}{2}$$ , $$a_{2 }$$= -$$\frac{3}{2}$$

(|) sinx.cosx =$$\frac{1}{2}$$ , (||) sinx.cosx = -$$\frac{3}{2}$$

(|) sinx= $$\frac{1}{2cosx}$$
$$sin^{2 }x + cos^{2 }x$$=1 ; $$\frac{1}{4 cos^{2 }x }$$ +$$cos^{2 }x$$=1 ; 4$$cos^{4 }x -4 cos^{2 }x$$+1 =0

$$(2 cos^{2 }x-1) ^{2 }$$=0 $$\Rightarrow$$ 2$$cos^{2 }x$$=1 ; $$cos^{2 }x$$ =$$\frac{1}{2}$$

cosx =$$\frac{ \sqrt{2} }{2}$$ $$\Rightarrow$$ $$x_{1 }$$=$$\frac{ \pi }{4}$$ +k$$\pi$$ ,cosx= -$$\frac{ \sqrt{2} }{2}$$ $$\Rightarrow$$ x=90$$^\circ$$+45$$^\circ$$+k$$\pi$$; x=$$\frac{ \pi }{2} + \frac{ \pi }{4} +k \pi$$; $$x_{2 }$$=$$\frac{3 \pi }{4} +k \pi$$

(||) sinx= -$$\frac{3}{2cosx}$$
$$sin^{2 }x+ cos^{2 }x =1$$ ; $$\frac{9}{4 cos^{2 }x }$$+$$cos^{2 }x$$=1 ;$$4cos^{4 }x -4 cos^{2 }x +9=0$$

$$4t^{2 }$$-4t+9=0 ; D=4-36<0 $$\Rightarrow$$ $$x_{3 }$$$$\in \varnothing$$
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