What is the general solution for 2cos^2(x)+cos^2(x)=0 ?

The general solution for 2cos^2(x)+cos^2(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin

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2cos^2(x)+cos^2(x)=0

2 cos^{2} (x) + cos^{2} (x) = 0

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (x) + cos^{2} (x) = 0

Simplify

2 cos^{2} (x) + cos^{2} (x) : 3 cos^{2} (x)

2 cos^{2} (x) + cos^{2} (x)

Add similar elements:

2 cos^{2} (x) + cos^{2} (x) = 3 cos^{2} (x)

= 3 cos^{2} (x)

3 cos^{2} (x) = 0

Divide both sides by

3

3 cos^{2} (x) = 0

Divide both sides by

3

3 cos^{2} (x) = 0

Divide both sides by

3

\frac{3 cos ^{2} ( x )}{3} = \frac{0}{3}

Simplify

cos^{2} (x) = 0

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (t) = \frac{1}{3}

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

\frac{9.03}{sin ( 4 0 ^{\circ} )} = \frac{10}{sin ( x )}

tan (θ) (sec (θ) - 1) = 0

tan (x) + sec (x) = 2

What is the general solution for 2cos^2(x)+cos^2(x)=0 ?
The general solution for 2cos^2(x)+cos^2(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin