What is the general solution for sin^2(x)+cos^2(2/5)=1 ?

The general solution for sin^2(x)+cos^2(2/5)=1 is x=0.4+2pin,x=pi-0.4+2pin,x=-0.4+2pin,x=pi+0.4+2pin

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sin^2(x)+cos^2(2/5)=1

sin^{2} (x) + cos^{2} (\frac{2}{5}) = 1

x = 0.4 + 2 πn, x = π - 0.4 + 2 πn, x = - 0.4 + 2 πn, x = π + 0.4 + 2 πn

Degrees

x = 22.91831 \dots^{\circ} + 36 0^{\circ} n, x = 157.08168 \dots^{\circ} + 36 0^{\circ} n, x = - 22.91831 \dots^{\circ} + 36 0^{\circ} n, x = 202.91831 \dots^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (x) + cos^{2} (\frac{2}{5}) = 1

Solve by substitution

sin^{2} (x) + cos^{2} (\frac{2}{5}) = 1

Let:

sin (x) = u

u^{2} + cos^{2} (\frac{2}{5}) = 1

u^{2} + cos^{2} (\frac{2}{5}) = 1 : u = 1 - cos^{2} (\frac{2}{5}), u = - 1 - cos^{2} (\frac{2}{5})

u^{2} + cos^{2} (\frac{2}{5}) = 1

Move

cos^{2} (\frac{2}{5})

to the right side

u^{2} + cos^{2} (\frac{2}{5}) = 1

Subtract

cos^{2} (\frac{2}{5})

from both sides

u^{2} + cos^{2} (\frac{2}{5}) - cos^{2} (\frac{2}{5}) = 1 - cos^{2} (\frac{2}{5})

Simplify

u^{2} = 1 - cos^{2} (\frac{2}{5})

u^{2} = 1 - cos^{2} (\frac{2}{5})

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1 - cos^{2} (\frac{2}{5}), u = - 1 - cos^{2} (\frac{2}{5})

Substitute back

u = sin (x)

sin (x) = 1 - cos^{2} (\frac{2}{5}), sin (x) = - 1 - cos^{2} (\frac{2}{5})

sin (x) = 1 - cos^{2} (\frac{2}{5}), sin (x) = - 1 - cos^{2} (\frac{2}{5})

sin (x) = 1 - cos^{2} (\frac{2}{5}) : x = arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn, x = π - arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn

sin (x) = 1 - cos^{2} (\frac{2}{5})

Apply trig inverse properties

sin (x) = 1 - cos^{2} (\frac{2}{5})

General solutions for

sin (x) = 1 - cos^{2} (\frac{2}{5})

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn, x = π - arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn

x = arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn, x = π - arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn

sin (x) = - 1 - cos^{2} (\frac{2}{5}) : x = arcsin (- 1 - cos^{2} (\frac{2}{5})) + 2 πn, x = π + arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn

sin (x) = - 1 - cos^{2} (\frac{2}{5})

Apply trig inverse properties

sin (x) = - 1 - cos^{2} (\frac{2}{5})

General solutions for

sin (x) = - 1 - cos^{2} (\frac{2}{5})

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- 1 - cos^{2} (\frac{2}{5})) + 2 πn, x = π + arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn

x = arcsin (- 1 - cos^{2} (\frac{2}{5})) + 2 πn, x = π + arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn

Combine all the solutions

x = arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn, x = π - arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn, x = arcsin (- 1 - cos^{2} (\frac{2}{5})) + 2 πn, x = π + arcsin (1 - cos^{2} (\frac{2}{5})) + 2 πn

Show solutions in decimal form

x = 0.4 + 2 πn, x = π - 0.4 + 2 πn, x = - 0.4 + 2 πn, x = π + 0.4 + 2 πn

3 \cdot cos (4 θ) = 0

5 tan (x) - 5 = 0

- 4 sin (x) + 3 cos (x) = 0

sin (x) - 1 = 0, sin (x) - 1 = 0

23 cos (2 x) = - 3

What is the general solution for sin^2(x)+cos^2(2/5)=1 ?
The general solution for sin^2(x)+cos^2(2/5)=1 is x=0.4+2pin,x=pi-0.4+2pin,x=-0.4+2pin,x=pi+0.4+2pin