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

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

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-1+sin^2(a)+cos^2(a)=cos^2(a)

- 1 + sin^{2} (a) + cos^{2} (a) = cos^{2} (a)

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

Degrees

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

Solution steps

- 1 + sin^{2} (a) + cos^{2} (a) = cos^{2} (a)

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= - 1 + 1

- 1 + 1 = cos^{2} (a)

Switch sides

cos^{2} (a) = - 1 + 1

Add/Subtract the numbers:

- 1 + 1 = 0

cos^{2} (a) = 0

Apply rule

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

cos (a) = 0

General solutions for

cos (a) = 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}

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

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

\frac{sin ( x )}{1 + cos ( x )} = 3 cot (\frac{x}{2})

sin (θ) = \frac{3}{2}, 0 \leq θ \leq 4 π

sin (2 x) = \frac{3}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

3 sin^{2} (x) + 8 cos^{2} (x) = 6

tan (θ) = \frac{2}{8}

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