What is the general solution for 2*cos^2(x)-sin^2(x)+1=0 ?

The general solution for 2*cos^2(x)-sin^2(x)+1=0 is x= pi/2+2pin,x=(3pi)/2+2pin

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2*cos^2(x)-sin^2(x)+1=0

2 \cdot cos^{2} (x) - sin^{2} (x) + 1 = 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) - sin^{2} (x) + 1 = 0

Rewrite using trig identities

1 - sin^{2} (x) + 2 cos^{2} (x)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

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

Simplify

= 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

1 = sinh (x)

15 cos^{2} (θ) + cos (θ) - 2 = 0

sin (θ) = \frac{9}{11}

- sin^{2} (x) + sin (x) = 0

3 sin^{2} (t) - 2 sin (t) - 1 = 0

What is the general solution for 2*cos^2(x)-sin^2(x)+1=0 ?
The general solution for 2*cos^2(x)-sin^2(x)+1=0 is x= pi/2+2pin,x=(3pi)/2+2pin