What is the general solution for cos^2(x)-sin^2(x)= 1/2 ,0<= x<= pi ?

The general solution for cos^2(x)-sin^2(x)= 1/2 ,0<= x<= pi is x= pi/6 ,x=(5pi)/6

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cos^2(x)-sin^2(x)= 1/2 ,0<= x<= pi

cos^{2} (x) - sin^{2} (x) = \frac{1}{2}, 0 \leq x \leq π

x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

cos^{2} (x) - sin^{2} (x) = \frac{1}{2}, 0 \leq x \leq π

Rewrite using trig identities

cos (2 x) = \frac{1}{2}

General solutions for

cos (2 x) = \frac{1}{2}

2 x = \frac{π}{3} + 2 πn, 2 x = \frac{5 π}{3} + 2 πn

Solve

2 x = \frac{π}{3} + 2 πn : x = \frac{π}{6} + πn

Solve

2 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{6} + πn

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

Solutions for the range

0 \leq x \leq π

x = \frac{π}{6}, x = \frac{5 π}{6}

sin^{2} (x) + 3 sin (x) - 2 = 0

cos (θ) = 0.9848

tan (x) = \frac{7}{4}

cos (θ) = \frac{2 5}{5}

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

What is the general solution for cos^2(x)-sin^2(x)= 1/2 ,0<= x<= pi ?
The general solution for cos^2(x)-sin^2(x)= 1/2 ,0<= x<= pi is x= pi/6 ,x=(5pi)/6